| What are the consequences of UD+ASSA? |
Rolf Nelson |
21/10/07 11:33 |
(Warning: This post assumes an familiarity with UD+ASSA and with the cosmological Measure Problem.) Observational Consequences: 1. Provides a possible explanation for the "Measure Problem" of why we shouldn't be "extremely surprised" to find we live in a lawful universe, rather than an extremely chaotic universe, or a homogeneous cloud of gas. 2. May help solve the Doomsday Argument in a finite universe, since you probably have at least a little more "measure" than a typical specific individual in the middle of a Galactic Empire, since you are "easier to find" with a small search algorithm than someone surrounded by enormous numbers of people. 3. For similar reasons, may help solve a variant of the Doomsday Argument where the universe is infinite. This variant DA asks, "if there's currently a Galactic Empire 10000 Hubble Volumes away with an immensely large number of people, why wasn't I born there instead of here?" 4. May help solve the Simulation Argument, again because a search algorithm to find a particular simulation among all the adjacent computations in a Galactic Empire is longer (and therefore, by UD +ASSA, has less measure) than a search algorithm to find you. 5. In basic UD+ASSA (on a typical Turing Machine), there is a probably a strict linear ordering corresponding to when the events at each point in spacetime were calculated; I would argue that we should expect to see evidence of this in our observations if basic UD+ASSA is true. However, we do not see any total ordering in the physical Universe; quite the reverse: we see a homogeneous, isotropic Universe. This is evidence (but not proof) that either UD+ASSA is completely wrong, or that if UD+ASSA is true, then it's run on something other than a typical linear Turing Machine. (However, if you still want use a different machine to solve the "Measure Problem", then feel free, but you first need to show that your non-Turing-machine variant still solves the "Measure Problem.") Decision Theory Consequences (Including Moral Consequences):
Every decision algorithm that I've ever seen is prey to paradoxes where the decision theory either crashes (fails to produce a decision), or requires an agent to do things that are bizarre, self- destructive, and evil. (If you like, substitute 'counter-intuitive' for 'bizarre, self-destructive, and evil.') For example: UD+ASSA, "Accepting the Simulation Argument", Utilitarianism without discounting, and Utilitarianism with time and space discounting all have places where they seem to fail. UD+ASSA, like the Simulation Argument, has the following additional problem: while some forms of Utilitarianism may only fail in hypothetical future situations (by which point maybe we'll have come up with a better theory), UD+ASSA seems to fail *right here and now*. That is, UD+ASSA, like the Simulation Argument, seems to call on you to do bizarre, self-destructive, and evil things today. An example that Yudowsky gave: you might spend resources on constructing a unique arrow pointing at yourself, in order to increase your measure by making it easier for a search algorithm to find you. Of course, I could solve the problem by deciding that I'd rather be self-destructive and evil than be inconsistent; then I could consider adopting UD+ASSA as a philosophy. But I think I'll pass on that option. :-) So, more work would have to be done the morality of UD+ASSA before any variant of UD+ASSA can becomes a realistically palatable part of a moral philosophy. -Rolf |
| Re: What are the consequences of UD+ASSA? |
Bruno Marchal |
24/10/07 06:19 |
Le 21-oct.-07, à 20:33, Rolf Nelson a écrit : > > (Warning: This post assumes an familiarity with UD+ASSA and with the > cosmological Measure Problem.) I am afraid you should say a little more on UD + ASSA. to make your points below clearer. I guess by UD you mean UDist (the universal distribution), but your remark remains a bit to fuzzy (at least for me) to comment. Of course I am not convinced by ASSA at the start, but still. The absence of recation of ASSA defenders is perhaps a symptom that you are not completely clear for them too?
Bruno http://iridia.ulb.ac.be/~marchal/
|
| Re: What are the consequences of UD+ASSA? |
Wei Dai |
24/10/07 18:25 |
Rolf Nelson wrote: > 1. Provides a possible explanation for the "Measure Problem" of why we > shouldn't be "extremely surprised" to find we live in a lawful > universe, rather than an extremely chaotic universe, or a homogeneous > cloud of gas. One thing I still don't understand, is in what sense exactly is the "Measure Problem" a problem? Why isn't it good enough to say that everything exists, therefore we (i.e. people living in a lawful universe) must exist, and therefore we shouldn't be surprised that we exist. If the "Measure Problem" is a problem, then why isn't there also an analogous "Lottery Problem" for people who have won the lottery? I admit that this "explanation" of why there is no problem doesn't seem satisfactory, but I also haven't been able to satisfactorily verbalize what is wrong with it. > Of course, I could solve the problem by deciding that I'd rather be > self-destructive and evil than be inconsistent; then I could consider > adopting UD+ASSA as a philosophy. But I think I'll pass on that > option. :-) I think our positions are pretty close on this issue, except that I do prefer to substitute 'counter-intuitive'. :-) The problem is, how can we be so certain that our intuitions are correct? > An example > that Yudowsky gave: you might spend resources on constructing a unique > arrow pointing at yourself, in order to increase your measure by > making it easier for a search algorithm to find you. While I no longer support UD+ASSA at this point (see my posts titled "against UD+ASSA"), I'm not sure this particular example is especially devastating. UD+ASSA perhaps implies an ethical theory in which all else being equal, you would prefer that there was a unique, easy to find arrow pointing at yourself. But it doesn't say that you should actually spend resources constructing it, since those resources might be better used in other ways, and it's not clear how much one's measure would actually be increased by such an arrow.
|
| Re: What are the consequences of UD+ASSA? |
Rolf Nelson |
24/10/07 20:14 |
On Oct 24, 9:25 pm, "Wei Dai" <wei...@weidai.com> wrote: > Rolf Nelson wrote: > > 1. Provides a possible explanation for the "Measure Problem" of why we > > shouldn't be "extremely surprised" to find we live in a lawful > > universe, rather than an extremely chaotic universe, or a homogeneous > > cloud of gas. > > One thing I still don't understand, is in what sense exactly is the "Measure > Problem" a problem? Why isn't it good enough to say that everything exists, > therefore we (i.e. people living in a lawful universe) must exist, and > therefore we shouldn't be surprised that we exist. If the "Measure Problem" > is a problem, then why isn't there also an analogous "Lottery Problem" for > people who have won the lottery? I don't have anything novel to say on the topic, but maybe if I restate the existing arguments, that'll help you expand on your counter-argument. The "Lottery Problem" would be a problem if I kept winning the lottery every day; I'd think something was fishy, and search for an explanation besides "blind chance", wouldn't you? Let's rank some classes of people, from chaotic (many rules) to lawful (few rules): 1. An infinite number of people live in "an infinite universe that obeys the Standard Model until November 1, 2007, and then adopts completely new laws of physics." If you live here, we predict that strange things will happen on November 1. 2. An infinite number of people live next-door in "an infinite universe that obeys the Standard Model through all of 2007, and maybe beyond." If you live here, expect nothing strange. 3. An infinite number of people live across the street in "a universe that looks like it obeys the Standard Model through November 1, 2007 because we are in the middle of a thermodynamic fluctuation, but the universe itself is extremely lawful, to the point where it's just a homogeneous gas with thermal fluctuations." We predict that strange things will happen on November 1. Your observations to date are consistent with all three models. What are the odds that you live in (2) but not (1) or (3)? Surely the answer is "extremely high", but how do we justify it *mathematically* (and philosophically)? If we can find mathematical solutions to satisfy this "Measure Problem", we can perhaps see what else that mathematical solution predicts, and test its predictions. Your UD+ASSA is the best solution I've seen so far, so I'm surprised there's not more interest in UD+ASSA (or some variant) as a "proto-science". >From the view of a potential scientific theory (rather than a philosophical "formalization of induction"), it's a *good* thing that it predicts "no oracles exist", because that is a falsifiable (though weak) prediction. |
| Re: What are the consequences of UD+ASSA? |
Bruno Marchal |
25/10/07 03:25 |
Le 25-oct.-07, à 03:25, Wei Dai a écrit : > > Rolf Nelson wrote: >> 1. Provides a possible explanation for the "Measure Problem" of why we >> shouldn't be "extremely surprised" to find we live in a lawful >> universe, rather than an extremely chaotic universe, or a homogeneous >> cloud of gas. > > One thing I still don't understand, is in what sense exactly is the > "Measure > Problem" a problem? Why isn't it good enough to say that everything > exists, > therefore we (i.e. people living in a lawful universe) must exist, and > therefore we shouldn't be surprised that we exist. If the "Measure > Problem" > is a problem, then why isn't there also an analogous "Lottery Problem" > for > people who have won the lottery? > > I admit that this "explanation" of why there is no problem doesn't seem > satisfactory, but I also haven't been able to satisfactorily verbalize > what > is wrong with it. Perhaps there can be a measure problem with the ASSA, or not. I have no idea because I think the ASSA idea, before having a measure problem, has a reference class problem. We don't know what is the set or class on which the measure can bear. If we say "observer", "observer-moment", "observer-life" etc... we have to define observer first, and each time this is done, it looks like I should be a bacteria instead of a human, or the measure cannot be well defined, or it presuppose a "physical world", etc. (see my old critics on ASSA, or on the Doomsday Argument. Now, with the COMP (and thus the RSSA), things change.The reference class is utterly well defined. For example, in the WM-duplication, it is the set {W,M}. In front of the UD, the reference class, although it is a non constructive object, it is, thanks to Church Thesis, a perfectly well defined mathematical object: it is the set of all states, going through your current state, generated by the DU. And the measure problem is made equivalent with the white rabbits problem (due to the existence of consistent but aberrant computations/histories (an history, I recall, is a computation as viewed from a first person perspective). If you disagress with this, it means you stop somewhere in between the first seven step of the 8-steps version of the UDA as in the slides http://iridia.ulb.ac.be/~marchal/publications/SANE2004Slide.pdf with explanations in http://iridia.ulb.ac.be/~marchal/publications/SANE2004MARCHAL.htm (html document), or http://iridia.ulb.ac.be/~marchal/publications/SANE2004MARCHAL.pdf (pdf document). I would be interested to know where. > >> Of course, I could solve the problem by deciding that I'd rather be >> self-destructive and evil than be inconsistent; then I could consider >> adopting UD+ASSA as a philosophy. But I think I'll pass on that >> option. :-) > > I think our positions are pretty close on this issue, except that I do > prefer to substitute 'counter-intuitive'. :-) The problem is, how can > we be > so certain that our intuitions are correct? > >> An example >> that Yudowsky gave: you might spend resources on constructing a unique >> arrow pointing at yourself, in order to increase your measure by >> making it easier for a search algorithm to find you. > > While I no longer support UD+ASSA at this point (see my posts titled > "against UD+ASSA"), I'm not sure this particular example is especially > devastating. UD+ASSA perhaps implies an ethical theory in which all > else > being equal, you would prefer that there was a unique, easy to find > arrow > pointing at yourself. But it doesn't say that you should actually spend > resources constructing it, since those resources might be better used > in > other ways, and it's not clear how much one's measure would actually be > increased by such an arrow.
I am not sure who "reads" that arrow, or even what *is* that arrow.
Bruno
http://iridia.ulb.ac.be/~marchal/
|
| Re: What are the consequences of UD+ASSA? |
Wei Dai |
25/10/07 04:59 |
Rolf Nelson wrote: > Your observations to date are consistent with all three models. What > are the odds that you live in (2) but not (1) or (3)? Surely the > answer is "extremely high", but how do we justify it *mathematically* > (and philosophically)? My current position is, forget the "odds". Let's say there is no odds, likelihood, probability, degrees of confidence, what have you, that I live in (2) but not (1) or (3). Instead, I'll consider myself as living in all of (1), (2), and (3), and whenever I make any decisions, I will consider the consequences of my choices on all of these universes. But the end result is that I'll still act *as if* I only live in (2) because I simply do not care very much about the consequences of my actions in (1) and (3). I don't care about (1) and (3) because those universes are too arbitrary or random, and I can defend that by pointing to their high algorithmic complexities. So this example does not seem to support the notion that the "Measure Problem" needs to be solved. > The "Lottery Problem" would be a problem if I kept winning the lottery > every day; I'd think something was fishy, and search for an > explanation besides "blind chance", wouldn't you? If I kept winning the lottery every day, I would have the following thoughts: There are two types of universe where I've won the lottery every day, those where there's a reason I've won (e.g., it's rigged to always let one person win) and those where there's no reason (i.e. I won them fair and square). I am living in universes of both types, but I care much more about those of the first type because they have lower algorithmic complexities. Therefore I should act as if I'm living in the first type of universe and try to find out what the reason is that I've won. But what if I've won the lottery only once? I'd still be tempted to ask "why did I win instead of someone else?" But the above rationale for searching for an answer doesn't work, because there is no simpler universe where a reason for my winning exists. The "Measure Problem" seems more like this situation. In both cases, there is no apparent rationale for asking "why", but we are tempted (or even compelled) to do so nevertheless.
|
| Re: What are the consequences of UD+ASSA? |
Tom Caylor |
25/10/07 09:22 |
On Oct 25, 3:25 am, Bruno Marchal <marc...@ulb.ac.be> wrote: > Le 25-oct.-07, à 03:25, Wei Dai a écrit : > > > Rolf Nelson wrote:
> >> An example > >> that Yudowsky gave: you might spend resources on constructing a unique > >> arrow pointing at yourself, in order to increase your measure by > >> making it easier for a search algorithm to find you. > > > While I no longer support UD+ASSA at this point (see my posts titled > > "against UD+ASSA"), I'm not sure this particular example is especially > > devastating. UD+ASSA perhaps implies an ethical theory in which all > > else > > being equal, you would prefer that there was a unique, easy to find > > arrow > > pointing at yourself. But it doesn't say that you should actually spend > > resources constructing it, since those resources might be better used > > in > > other ways, and it's not clear how much one's measure would actually be > > increased by such an arrow. > > I am not sure who "reads" that arrow, or even what *is* that arrow. > > Bruno > How about SAI (Super Intelligence)? Or God? Seriously, of course. The problem with generic SAI is the one you brought up: how do you know the SAI is good? This problem does not exist with a good God. Also the problem of what is the arrow, how do you make it, does not exist with the Christian God, since the Christian God (and no other one) made the arrow himself. Tom |
| Re: What are the consequences of UD+ASSA? |
Bruno Marchal |
26/10/07 03:24 |
Le 25-oct.-07, à 18:22, Tom Caylor a écrit : > How about SAI (Super Intelligence)? Or God? Seriously, of course. > The problem with generic SAI is the one you brought up: how do you > know the SAI is good? This problem does not exist with a good God. > Also the problem of what is the arrow, how do you make it, does not > exist with the Christian God, since the Christian God (and no other > one) made the arrow himself. Hmmm.... It seems to me you are quite quick here. Especially after reading Vance novels, as linked by Marc. Is God good? Well, according to Plato, accepting the rather natural "theological" interpretation of the Parmenides (like Plotinus), there is a sense to say that God is "good", but probably not in the Christian sense (if that can be made precise). Indeed, Plato's God is just Truth. And Truth is not good as such, but the awareness of truth, or simply the search of truth, is, for a Platonist, a prerequisite for the *possible* development of goodness. Truth is necessary for justice, and justice is necessary for goodness. That's the idea. It makes knowledge (and thus truth) a good thing, in principle. But Vance's novel rises a doubt. Actually, that doubt can rise through the reading of the first Pythagorean writings, which insist so much on hiding their knowledge to the non-initiated people, making them secret. (according to the legend, their kill a disciple who dares to make public the discovery of the irrationality of the square root of 2). Maimonides also, in his "Guide for the perplexed" insists that fundamental knowledge has to be reserved for the initiated or the elite people. Fundamentally I don't know. I know a lot of particular case where knowledge can be bad. But this happens always in "human, too much human" practical circumstances, like during war, illness, etc. (it is not good that your enemies *knows* where are your missiles; it is not good to tell a bad new to some old dying people, etc. But this never concerns fundamental truth. I guess it *is* a question of faith. Of course, something like complete knowledge, would be bad, making life without any purpose (at least it is natural to fear that), but in this case both lobianity, and well, may be things like Christianity, remind us about our finiteness and about the fact that complete knowledge is inconsistent (even for Gods, but not for the Unnameable, making it above thinking (something Plotinus understood, but I am not sure Christians, following here Aristotle theology, take this seriously into account but then they do have confuse temporal and spiritual power isn't it?). Now, Tom, to come back to the present thread, i.e. Wei Dai's question on the meaning of the measure problem with respect to the ASSA philosophy, frankly I am not sure that saying that God is responsible for the indexical "arrow" will put light. It looks a bit like closing even the possibility of progressing, given that God can hardly be invoked in any attempt to scientifically explains something (cf "scientifically" means based on a clear and doubtable (if not refutable) theory). So you would have to elaborate, but as we have already discussed, to use God here would mean that you do have a doubtable and clear theory of God. OK if you are using lobian theology (which is cristal clear I think), but which cannot be related so easily with any human religion without much work on both human and machine and comp, etc. We would quickly been led to propositions far more difficult, not to say controversial, than Wei Dai's original question. Of course, here, those who take the primacy of a physical universe for granted, somehow, makes the same mistake than those who take God or a God for granted. Such moves hide the questions through incommunicable (perhaps even false) "certitude". Bruno http://iridia.ulb.ac.be/~marchal/
|
| Re: What are the consequences of UD+ASSA? |
Rolf Nelson |
26/10/07 05:19 |
In standard decision theory, "odds" (subjective probabilities) are separated from utilities. Is "how much you care about the consequences of your actions" isomorphic to "odds", or is there some subtlety I'm missing here? One thing unclear is whether you're advocating "moral relativism", or whether you simply want an "escape clause" in your formal decision theory so that if you don't like what your decision theory tells you to do, you can alter your decision theory on the spot on a case-by- case basis. Wei Dai wrote: > I simply do not care very much about the consequences of my actions in (1) and (3).
|
| Re: What are the consequences of UD+ASSA? |
Brent |
26/10/07 11:18 |
Bruno Marchal wrote: ... > > Is God good? Well, according to Plato, accepting the rather natural > "theological" interpretation of the Parmenides (like Plotinus), there > is a sense to say that God is "good", but probably not in the Christian > sense (if that can be made precise). Indeed, Plato's God is just Truth. > And Truth is not good as such, but the awareness of truth, or simply > the search of truth, is, for a Platonist, a prerequisite for the > *possible* development of goodness. > Truth is necessary for justice, and justice is necessary for goodness. > That's the idea. It makes knowledge (and thus truth) a good thing, in > principle. > But Vance's novel rises a doubt. Actually, that doubt can rise through > the reading of the first Pythagorean writings, which insist so much on > hiding their knowledge to the non-initiated people, making them secret. > (according to the legend, their kill a disciple who dares to make > public the discovery of the irrationality of the square root of 2). > Maimonides also, in his "Guide for the perplexed" insists that > fundamental knowledge has to be reserved for the initiated or the elite > people. > > Fundamentally I don't know. I know a lot of particular case where > knowledge can be bad. But this happens always in "human, too much > human" practical circumstances, like during war, illness, etc. (it is > not good that your enemies *knows* where are your missiles; it is not > good to tell a bad new to some old dying people, etc. But this never > concerns fundamental truth.
But what truth is "fundamental"? Quantum gravity seems like an esoteric game to most people and so you can say anything you want about it without any ethical implications. But when quantum gravity seems to provide a non-supernatural cosmogony, religions are threatened and suddenly it's like bad news to a dying man (and we're all dying). Coincidentally, James Watson has just lost his job because he said some things that, while narrowly true, support a racist view of Africa. Were they "fundamental" or does "fundamental" = "of no import in society"? Brent Meeker |
| Re: What are the consequences of UD+ASSA? |
Wei Dai |
26/10/07 14:28 |
Rolf Nelson wrote: > In standard decision theory, "odds" (subjective probabilities) are > separated from utilities. Is "how much you care about the consequences > of your actions" isomorphic to "odds", or is there some subtlety I'm > missing here? Your question shows that someone finally understand what I've been trying to say, I think. "how much you care about the consequences of your actions" is almost isomorphic to "odds", except that I've found a couple of cases where thinking in terms of the former works (i.e. delivers intuitive results) whereas the latter doesn't. The first I described in "against UD+ASSA, part 1" at http://groups.google.com/group/everything-list/browse_frm/thread/dd21cbec7063215b. The second one is, what if your preferences for two universes are not independent? For example, suppose you have the following preferences, from most preferred to least preferred: 1) eat an apple in universe A and eat an orange in universe B 2) eat an orange in universe A and eat an apple in universe B 3) eat an apple in both universes 4) eat an orange in both universes I don't see why this kind of preference must be irrational if you believe that both A and B exists. But in standard decision theory, this kind of preference is not allowed. To put it more generally, thinking in terms of "how much you care about the consequences of your actions" *allows* you to have an overall preference about A and B that can be expressed as an expected utility: P(A) * U(A) + P(B) * U(B) since P(A) and P(B) can denote how much you care about universes A and B, but it doesn't *force* you to have a preference of this form. Standard decision theory does force you to. > One thing unclear is whether you're advocating "moral relativism", or > whether you simply want an "escape clause" in your formal decision > theory so that if you don't like what your decision theory tells you > to do, you can alter your decision theory on the spot on a case-by- > case basis. That's a very good question. I think if someone were to show me an objective decision procedure that actually makes sense, I think I would give up "moral relativism". But in the mean time, I don't see how to avoid these counterintuitive implications without it.
|
| Re: What are the consequences of UD+ASSA? |
Rolf Nelson |
27/10/07 09:33 |
Wei, your examples are convincing, although other decision models have similar problems. If your two examples were the only problems that UDASSA had, I would have few qualms about adopting it over the other decision models I've seen. Note that even if you adopt a decision model, you still in practice (as a human being) can keep an all- purpose "escape hatch" where you can go against your formal model if there are edge cases where you dislike its results. In other words, I would prioritize "UDASSA doesn't yet make many falsifiable predictions" and "We don't see a total ordering of points in spacetime, so UDASSA probably doesn't run on a typical Turing Machine" as larger problems. But sure, if UDASSA can be improved to solve the morality edge-cases that you gave, I'm all for the improvements. As far as our observations of the Universe, I don't quite follow: how can you go from "in terms of morality, probability is imperfect" to "there's no such thing as probability, therefore there's no measure problem?" |
| Re: What are the consequences of UD+ASSA? |
Rolf Nelson |
27/10/07 10:18 |
> To put it more generally, thinking in terms of "how much you care about the > consequences of your actions" *allows* you to have an overall preference > about A and B that can be expressed as an expected utility: > > P(A) * U(A) + P(B) * U(B) > > since P(A) and P(B) can denote how much you care about universes A and B, > but it doesn't *force* you to have a preference of this form. Standard > decision theory does force you to. True. So how would an alternative scheme work, formally? Perhaps utility can be formally based on the "Measure" of "Qualia" (observer moments). If you have a halting oracle, certain knowledge of a Universal Prior, and infinite cognitive resources, you can choose your action to maximize a utility function U(X); X is the sequence M(Q1), M(Q2), ..., where the measures of all possible Qualia are enumerated. In the typical case of everyday life decisions in 2007, M would often reduce to an objective probability oP; and U(X) = U(M(Q1), M(Q2), ...) maybe has an affine (in other words, a decision-theory-order- preserving) transformation, for a typical 2007 human, to some function U(how good life is expected to be for earthly observer O1, how good life is expected to be for earthly observer O2, ...), (pretending for now that you don't have any way of altering the "total measure" taken up by a human being.) "How good life is expected to be for observer O1" in turn perhaps reduces, in typical life, to oP(O1 experiences Q1) * (desirableness of Q1) + oP(O1 experiences Q2) * (desirableness of Q2) + ... But now we have to say that no one actually has infinite cognitive resources, let alone a halting Oracle. So, we probably still want a "logical probability" lP to deal with things like "To what extent do I currently believe that the Riemann Hypothesis is true." So you can't choose an action to maximize U directly, instead you want to maximize the expected utility, by maximizing the following: lP(X1) * U(X1) + lP(X2) * U(X2) + ... Humans would perceive, as "subjective probability", a combination of the Measure-based "objective probability" and the logic-based "logical probability". Clear as mud, I'm sure. Plus the odds are that I got something wrong in the details. But that's my take on it, anyway. |
| Re: What are the consequences of UD+ASSA? |
Günther Greindl |
27/10/07 14:43 |
Hi all, > One thing I still don't understand, is in what sense exactly is the "Measure > Problem" a problem? Why isn't it good enough to say that everything exists, > therefore we (i.e. people living in a lawful universe) must exist, and > therefore we shouldn't be surprised that we exist. If the "Measure Problem" > is a problem, then why isn't there also an analogous "Lottery Problem" for > people who have won the lottery? thank you Wei Dei, I have expressed something similar concerning the Doomsday Argument which has the same reasoning flaw.
You can't reason about probabilities "inside" the system and be surprised that you are in "location" A or B. Example: 1) If I draw from an urn with 1 Million white balls and 1 black ball, I should be pretty surprised if I draw the black one. 2) If I am a black ball in an urn (same distribution as above) and I only become conscious if I am drawn and I suddenly "wake up" to find myself drawn, I shouldn't be surprised at all - my being drawn was a condition for being a perceptive being. I think a mixing up of these two viewpoints underly much of "measure problem", doomsday and other arguments of the same sort. Regards, Günther -- Günther Greindl Department of Philosophy of Science University of Vienna guenther...@univie.ac.at http://www.univie.ac.at/Wissenschaftstheorie/
Blog: http://dao.complexitystudies.org/ Site: http://www.complexitystudies.org
|
| Re: What are the consequences of UD+ASSA? |
Brent |
27/10/07 15:11 |
Günther Greindl wrote: > Hi all, > >> One thing I still don't understand, is in what sense exactly is the "Measure >> Problem" a problem? Why isn't it good enough to say that everything exists, >> therefore we (i.e. people living in a lawful universe) must exist, and >> therefore we shouldn't be surprised that we exist. If the "Measure Problem" >> is a problem, then why isn't there also an analogous "Lottery Problem" for >> people who have won the lottery? That's a good argument assuming some laws of physics. But as I understood it, the "measure problem" was to explain the law-like evolution of the universe as a opposed to a chaotic/random/white-rabbit universe. Is it your interpretation that, among all possible worlds, somebody has to live in law-like ones; so it might as well be us? Brent Meeker > > > thank you Wei Dei, I have expressed something similar concerning the > Doomsday Argument which has the same reasoning flaw. > > You can't reason about probabilities "inside" the system and be > surprised that you are in "location" A or B. > > Example: > > 1) If I draw from an urn with 1 Million white balls and 1 black ball, I > should be pretty surprised if I draw the black one. > > 2) If I am a black ball in an urn (same distribution as above) and I > only become conscious if I am drawn and I suddenly "wake up" to find > myself drawn, I shouldn't be surprised at all - my being drawn was a > condition for being a perceptive being. > > I think a mixing up of these two viewpoints underly much of "measure > problem", doomsday and other arguments of the same sort. > > Regards, > Günther > >
|
| Re: What are the consequences of UD+ASSA? |
Wei Dai |
27/10/07 19:59 |
Rolf Nelson wrote: > Wei, your examples are convincing, although other decision models have > similar problems. If your two examples were the only problems that > UDASSA had, I would have few qualms about adopting it over the other > decision models I've seen. Note that even if you adopt a decision > model, you still in practice (as a human being) can keep an all- > purpose "escape hatch" where you can go against your formal model if > there are edge cases where you dislike its results. For me, this line of thought started with the question "what does probability mean if everything exists?" (Actually, before that I had thought about "what does probability mean if brain copying is possible?") I've entertained many different possible answers. I looked at decision theories not because I'm looking for a decision procedure to adopt, but because that is one way probability is interpreted and justified. I'm actually more interested in the philosophical issues rather than the practical ones. Besides, if you program a decision procedure into an AI, it had better be flawless because there may be no "escape hatches". > In other words, I would prioritize "UDASSA doesn't yet make many > falsifiable predictions" and "We don't see a total ordering of points > in spacetime, so UDASSA probably doesn't run on a typical Turing > Machine" as larger problems. But sure, if UDASSA can be improved to > solve the morality edge-cases that you gave, I'm all for the > improvements. I consider UD+ASSA to be a theory of how people reason, or how they ought to reason, and as such, it does make falsifiable predictions. In fact, as I showed in several examples, the predictions have been falsified. About your comment "We don't see a total ordering of points in spacetime, so UDASSA probably doesn't run on a typical Turing Machine". I don't follow your reasoning here as to why UD+ASSA+typical TM implies that we should see a total ordering of points in spacetime. Isn't it possible that such an ordering exists internal to the TM's program, but it's not visible to the people inside the universe that the TM simulates? > As far as our observations of the Universe, I don't quite follow: how > can you go from "in terms of morality, probability is imperfect" to > "there's no such thing as probability, therefore there's no measure > problem?" My reasoning goes like this: 1. We need to reinterpret probability, from "subjective degree of belief" to "how much do I care about something" in order to fix counterintuitive implications of decision theory. 2. Once we do that, we no longer seem to have a solution to the "measure problem". 3. Let's look closer at the nature of the problem. It seems to consist of two parts: (A) Why am I living in an apparently lawful universe? (B) Why should I expect the future to continue to be lawful? 4. I think (B) is the easier question, and I answered it in a previous post in this thread. (A) is more problematic, but my tentative answer is that, as Brent Meeker stated it, "among all possible worlds, somebody has to live in law-like ones; so it might as well be us." I'm out of time today, and will respond to your other post tomorrow.
|
| Re: What are the consequences of UD+ASSA? |
Wei Dai |
27/10/07 20:00 |
Brent Meeker wrote: > That's a good argument assuming some laws of physics. But as I understood > it, the "measure problem" was to explain the law-like evolution of the > universe as a opposed to a chaotic/random/white-rabbit universe. Is it > your interpretation that, among all possible worlds, somebody has to live > in law-like ones; so it might as well be us? Yes. See my other post today.
|
| Re: What are the consequences of UD+ASSA? |
Rolf Nelson |
28/10/07 08:25 |
On Oct 25, 7:59 am, "Wei Dai" <wei...@weidai.com> wrote: > I don't care > about (1) and (3) because those universes are too arbitrary or random, and I > can defend that by pointing to their high algorithmic complexities. In (3) the universe doesn't have a high aIgorithmic complexity. Any theory that just says "we only care about universes with low algorithmic complexity" leads to (3) (assuming that, by "the universe", you have the usual meaning of "that vast space we seem to live in" rather than "my immediate perceptions".) The specific reason I like UDASSA is because it gives you a framework for saying, "the universe, plus my index in the universe, has a low algorithmic complexity." |
| Re: What are the consequences of UD+ASSA? |
Rolf Nelson |
28/10/07 08:49 |
> About your comment "We don't see a total ordering of points in spacetime, so > UDASSA probably doesn't run on a typical Turing Machine". I don't follow > your reasoning here as to why UD+ASSA+typical TM implies that we should see > a total ordering of points in spacetime. Isn't it possible that such an > ordering exists internal to the TM's program, but it's not visible to the > people inside the universe that the TM simulates? It definitely is possible, my only point is that the fact that most UTM program outputs don't have an easily-observed homogeneous and isotropic n-dimensional space in their output, *may* be Bayesian evidence against the plain UDASSA. So if we consider 3 hypotheses: 1. plain UDASSA 2. UDASSA variants, such as the set-theory UDASSA you mentioned, or a UDASSA on a UTM that can atomically implement higher-level operations like "multiply two complex numbers to infinite precision" and "apply an operation uniformly to an infinite manifold". 3. something else Then the lack of ordering that we see probably gives me a "Bayesian Shift" from (1) to (2) or (3). However, to demonstrate would probably be difficult, and if we had something powerful enough to do this, we might have a theory that allows UDASSA to make novel predictions about the observed Universe. |
| Re: What are the consequences of UD+ASSA? |
Rolf Nelson |
28/10/07 09:08 |
> However, to demonstrate would probably > be difficult, and if we had something powerful enough to do this, we > might have a theory that allows UDASSA to make novel predictions about > the observed Universe. To give examples of how hard this is: 1. What is the probability that our Universe has existed since the Big Bang, but will abruptly end tomorrow? There have been about 2^16 days since the Big Bang, so we can get a lower bound of probability in UDASSA with 1 / 2^((length of a binary program that runs a Universe for x subjective time, then halts) + (about 16 bits)). I don't know how to program any of the basic TM's, and can't personally estimate of the complexity of the first term. And this is just to get an lower bound, the actual probability is probably much higher. 2. Take a real-world example, like the Pioneer Anomaly; does "new laws of physics caused the Pioneer Anomaly" have a higher or lower complexity than "there is a mundane explanation for the Pioneer Anomaly"? Good luck! On the plus side, one wouldn't have to solve every problem to make UDASSA into a science; one would just have to solve (successfully predict) a handful of novel problems (that aren't solvable by other methods) to demonstrate that is true and useful. |
| Re: What are the consequences of UD+ASSA? |
Wei Dai |
28/10/07 18:44 |
Rolf Nelson wrote: > On Oct 25, 7:59 am, "Wei Dai" <wei...@weidai.com> wrote: >> I don't care >> about (1) and (3) because those universes are too arbitrary or random, >> and I >> can defend that by pointing to their high algorithmic complexities. > > In (3) the universe doesn't have a high aIgorithmic complexity. I should have said that in (3) our decisions don't have any consequences, so we disregard them even if we do care what happens in them. The end result is the same: I'll act as if I only live in (2). From your post yesterday: > True. So how would an alternative scheme work, formally? Perhaps > utility can be formally based on the "Measure" of "Qualia" (observer > moments).
This is one of the possibilities I had considered and rejected, because it also leads to counterintuitive consequences. For example, suppose someone gives your the following offer: I will throw a fair coin. If the coin lands heads up, you will be instantaneously vaporized. If it lands tails up, I will exactly double your measure (say by creating a copy of your brain and continuously keeping it synchronized). Given your "measure of qualia"-based formalization of utility, and assuming that you're selfish so that you're only interested in the measure of the qualia of your own future selves, you'd have to be indifferent between accepting this offer and not accepting it. Instead, here's my current approach for a formalization of decision theory. Let a set S be the description of an agent's knowledge of the multiverse. For example, for a Tegmarkian version of the multiverse, elements of S have the form (s, t) where s is a statement of second-order logic, and t is either "true" or "false". For simplicity, assume that the decision-making agent is logically omniscient, which means he knows the truth value of all statements of second-order logic, except those that depend on his own decisions. We'll say that he prefers choice A to choice B if and only if he prefers S U C(E,A) to S U C(E,B), where U is the union operator, C(x,y) is the logical consequences of everyone having qualia x deciding to do y, and E consists of all of his own memories and observations. In this most basic version, there is not even a notion of "how much one cares about a universe". I'm relatively confident that it doesn't lead to any counterintuitive implications, but that's mainly because it is too weak to lead to any kind of implications at all. So how do we explain what probability is, and why the concept has been so useful? Well, let's consider an agent who happens to have preferences of a special form. It so happens that for him, the multiverse can be divided into several "regions", the descriptions of which will be denoted S_1, S_2, S_3, etc., such that S_1 U S_2 U S_3 ... = S and his preferences over the whole multiverse can be expressed as a linear combination of his preferences over those "regions". That means, there exists functions P(.) and U(.) such that he prefers the multiverse S to the multiverse T if and only if P(S_1)*U(S_1) + P(S_2)*U(S_2) + P(S_3)*U(S_3) + ... > P(T_1)*U(T_1) + P(T_2)*U(T_2) + P(T_3)*U(T_3) ... I haven't worked out all of the details of this formalism, but I hope you can see where I'm going with this...
|
| Re: What are the consequences of UD+ASSA? |
Bruno Marchal |
29/10/07 07:28 |
How do the UDASSA, or the UDISTASSA, people take the difference between first person and third person into account? Do they? With the RSSA (through the use of the UD) it should be clear that THIRD person determinism and computability entails FIRST person indeterminacy and "observable non computability" (like what we can "see" when preparing many particles in the state 1/sqrt(2)(up+down) and looking them in the base {up, down}. Bruno http://iridia.ulb.ac.be/~marchal/
|
| Re: What are the consequences of UD+ASSA? |
Bruno Marchal |
30/10/07 07:28 |
Le 26-oct.-07, à 20:18, Brent Meeker a écrit : OK. I should not have talk about "fundamental truth", but about "fundamental question". By which I mean "where do I come from?", "what can we hope", "what is the nature of matter", "is life a dream?" etc. None of those question are really addressed by today's science which focuses on the physical aspect of reality without really tackling seriously (in the doubting way) its metaphysical aspect and its psycho or theo logical aspect. > Quantum gravity seems like an esoteric game to most people and so you > can say anything you want about it without any ethical implications. > But when quantum gravity seems to provide a non-supernatural > cosmogony, religions are threatened and suddenly it's like bad news to > a dying man (and we're all dying).
If a religion is threatened by science, it means it is build on bad faith. At least with comp science has to be a part of theology. If theology does not extend science, it means it is wrong. Now sometimes some scientist talk like if they were priest, and that is two times more wrong than usual priest talk. Religion, like science, is threatened by bad science, and even more by bad religion (religion based on "blind" faith or authoritative argument). Now, don't tell me that a theory like quantum gravity provides a non supernatural cosmogony given that quantum gravity study quantum gravity and perhaps the physical universe, but not the mind, nor the soul, the person or consciousness. It is not its subject a priori. To look at quantum gravity as a cosmogony is a confusion between subject like physics and theology. This threatens theology, because without making some very strong physicalist assumption, which are incompatible with the mechanist thesis, it consists to make physics a religion without saying! This is just dishonest. Such a physical universe is worst than a white male God, because it looks scientific (unlike the male God), but it isn't.
It looks that in winter, people forget all about the 1/3 distinction. In Quantum gravity this 1/3 distinction is a bit hidden. You have to postulate comp, and thus Everett before. (Well, as you know, you have to derive Everett but it is not the point here). > > Coincidentally, James Watson has just lost his job because he said > some things that, while narrowly true, support a racist view of > Africa. Were they "fundamental" or does "fundamental" = "of no > import in society"?
I love Watson because I discover "the math of computer science" by myself in his book "Molecular Biology of the Gene". This book has played a so big role in my youth that I have been using for years the word "Watson" as a synonym of "Bible". But J. Watson has become the worst materialist I have ever heard about. According to a talk I have followed some years ago (I should search for the reference) Watson seems to believe only in ATOMS". Someone told him "Surely you believe in molecules M. Watson". And James Watson would have answered: "No, I don't, there are only atoms!". Weird .... By "fundamental" I really mean the same as in "fundamental science". Unless in company of theological hypothesis, it has no more impact on society other than its technical products. Einstein discovered the relation between matter and energy only through a deep motivation for fundamental question: what is matter, what is the nature of light, how could resemble the universe when seen by a photon, etc. But of course, its questioning led him to the discovery of precise and refutable empirical statements, most of them did have incredible impacts of our live today. Bruno http://iridia.ulb.ac.be/~marchal/
|
| Re: What are the consequences of UD+ASSA? |
Rolf Nelson |
30/10/07 17:03 |
> > In (3) the universe doesn't have a high aIgorithmic complexity. > > I should have said that in (3) our decisions don't have any consequences, so > we disregard them even if we do care what happens in them. The end result is > the same: I'll act as if I only live in (2). In the (3) I gave, you're indexed so that the thermal fluctuation doesn't dissolve until November 1, so your actions still have consequences. > I will throw a fair coin. If the coin lands heads up, you will be > instantaneously vaporized. If it lands tails up, I will exactly double your > measure (say by creating a copy of your brain and continuously keeping it > synchronized). This is one of a larger class of problems related to volition, and the coupling of my qualia to an external reality, that I don't currently have an answer for. I want to live on in the current Universe, I don't to die and have a duplicate of myself created in a different Universe. I want to eat a real ice cream cone, I don't want you to stimulate my neurons to make me imagine I'm eating an ice cream cone. I would argue that a world where I can interact with real people is, in some sense, better than a world where I interact with imaginary people who I believe are real. > Well, let's consider an agent who happens to have preferences of a special > form. It so happens that for him, the multiverse can be divided into several > "regions", the descriptions of which will be denoted S_1, S_2, S_3, etc., > such that S_1 U S_2 U S_3 ... = S and his preferences over the whole > multiverse can be expressed as a linear combination of his preferences over > those "regions". That means, there exists functions P(.) and U(.) such that > he prefers the multiverse S to the multiverse T if and only if > > P(S_1)*U(S_1) + P(S_2)*U(S_2) + P(S_3)*U(S_3) + ... > > > P(T_1)*U(T_1) + P(T_2)*U(T_2) + P(T_3)*U(T_3) ... > > I haven't worked out all of the details of this formalism, but I hope you > can see where I'm going with this... You have a general model, which can encompass classical decision theory, but can also encompass other models as well. It's not immediately clear to me what benefit, if any, we get from such a general model. |
| Re: What are the consequences of UD+ASSA? |
Wei Dai |
30/10/07 19:28 |
Rolf Nelson wrote: > In the (3) I gave, you're indexed so that the thermal fluctuation > doesn't dissolve until November 1, so your actions still have > consequences. Still not a problem: the space-time region that I can affect in (3) is too small (i.e., its measure is too small, complexity too large) for me to care much about the consequences of my actions on it. > This is one of a larger class of problems related to volition, and the > coupling of my qualia to an external reality, that I don't currently > have an answer for. I want to live on in the current Universe, I don't > to die and have a duplicate of myself created in a different Universe. > I want to eat a real ice cream cone, I don't want you to stimulate my > neurons to make me imagine I'm eating an ice cream cone. I would argue > that a world where I can interact with real people is, in some sense, > better than a world where I interact with imaginary people who I > believe are real. To me, these examples show that we do not care just about qualia, but also about attributes and features of the multiverse that can not be classified as qualia, and therefore we should rule out decision theories that cannot incorporate preferences over non-qualia. > You have a general model, which can encompass classical decision > theory, but can also encompass other models as well. It's not > immediately clear to me what benefit, if any, we get from such a > general model. Fair question. I'll summarize: 1. We are forced into considering such a general model because we don't have a more specific one that doesn't lead to counterintuitive implications. 2. It shows us what probabilities really are. For someone whose preferences over the multiverse can be expressed as a linear combination of preferences over regions of the multiverse, a probability function can be interpreted as a representation of how much he cares about each region. I would argue that most of us in fact have preferences of this form, at least approximately, which explains why probability theory has been useful for us. 3. It gives us a useful framework for considering anthropic reasoning problems such as the Doomsday Argument and the Simulation Argument. We can now recast these questions into "Do we prefer a multiverse where people in our situation act as if doom is near?" and "Do we prefer a multiverse where people in our situation act as if they are in simulations?" I argue that its easier for us to consider these questions in this form. 4. For someone on a practical mission to write an AI that makes sensible decisions, perhaps the model can serve as a starting point and as illustration of how far away we still are from that goal.
|
| Re: What are the consequences of UD+ASSA? |
Brent |
30/10/07 19:53 |
Wei Dai wrote: > Rolf Nelson wrote: >> In the (3) I gave, you're indexed so that the thermal fluctuation >> doesn't dissolve until November 1, so your actions still have >> consequences. > > Still not a problem: the space-time region that I can affect in (3) is too > small (i.e., its measure is too small, complexity too large) for me to care > much about the consequences of my actions on it. > >> This is one of a larger class of problems related to volition, and the >> coupling of my qualia to an external reality, that I don't currently >> have an answer for. I want to live on in the current Universe, I don't >> to die and have a duplicate of myself created in a different Universe. >> I want to eat a real ice cream cone, I don't want you to stimulate my >> neurons to make me imagine I'm eating an ice cream cone. I would argue >> that a world where I can interact with real people is, in some sense, >> better than a world where I interact with imaginary people who I >> believe are real. > > To me, these examples show that we do not care just about qualia, but also > about attributes and features of the multiverse that can not be classified > as qualia, and therefore we should rule out decision theories that cannot > incorporate preferences over non-qualia. > >> You have a general model, which can encompass classical decision >> theory, but can also encompass other models as well. It's not >> immediately clear to me what benefit, if any, we get from such a >> general model. > > Fair question. I'll summarize: > > 1. We are forced into considering such a general model because we don't have > a more specific one that doesn't lead to counterintuitive implications. > > 2. It shows us what probabilities really are. For someone whose preferences > over the multiverse can be expressed as a linear combination of preferences > over regions of the multiverse, a probability function can be interpreted as > a representation of how much he cares about each region. I would argue that > most of us in fact have preferences of this form, at least approximately, > which explains why probability theory has been useful for us. This seems to just reverse the decision theoretic meaning of probability. Usually one cares more about probables outcome and ignores the very improbable ones. For example I prefer a region in which I'm rich, handsome, and loved by all beautiful women - but I don't assign much probability to it. > > 3. It gives us a useful framework for considering anthropic reasoning > problems such as the Doomsday Argument and the Simulation Argument. We can > now recast these questions into "Do we prefer a multiverse where people in > our situation act as if doom is near?" and "Do we prefer a multiverse where > people in our situation act as if they are in simulations?" I argue that its > easier for us to consider these questions in this form. But it seems the answer might depend on whether the premise were true - which makes the problem harder. Brent Meeker > > 4. For someone on a practical mission to write an AI that makes sensible > decisions, perhaps the model can serve as a starting point and as > illustration of how far away we still are from that goal. > > > > > > > > |
| Re: What are the consequences of UD+ASSA? |
marc.geddes |
30/10/07 22:32 |
On Oct 31, 3:28 pm, "Wei Dai" <wei...@weidai.com> wrote: > > 4. For someone on a practical mission to write an AI that makes sensible > decisions, perhaps the model can serve as a starting point and as > illustration of how far away we still are from that goal. Heh. Yes, very interesting indeed. But a huge body of knowledge and a great deal of smartness is needed to even begin to grasp all that stuff ;) As regards AI I gotta wonder whether that 'Decision Theory' stuff is really 'the heart of the matter' - perhaps its the wrong level of abstraction for the problem. That is it say, it would be great if the AI could work out all the decision theory for itself, rather than having us trying to program it in (and probably failing miserably). Certainly, I'm sure as hell not smart enough to come up with a working model of decisions. So, rather than trying to do the impossible, better to search for a higher level of abstraction. Look for the answers in communication theory/ontology, rather than decision theory. Decision theory would be derivative of an effective ontology - that saves me the bother of trying to work it out ;) |
| Re: What are the consequences of UD+ASSA? |
Brent |
30/10/07 23:40 |
Decisions require some value structure. To get values from an ontology you'd have to get around the Naturalistic fallacy. Brent Meeker
|
| Re: What are the consequences of UD+ASSA? |
marc.geddes |
31/10/07 00:05 |
On Oct 31, 7:40 pm, Brent Meeker <meeke...@dslextreme.com> wrote: > > Decisions require some value structure. To get values from an ontology you'd have to get around the Naturalistic fallacy. >
> Brent Meeker- Hide quoted text - > > - Show quoted text - Decision theory has this same problem. Decision theory doesn't require values. The preferences (values) are plugged in from outside the theory. Decision theory is merely a way of computing the best way to achieve the desired outcomes. It doesn't say what we should desire though. Decision theory is too hard for me and too complex. What I'm suspecting is that it's not the final word. I'm looking for a higher level theory capable of deriving the results in decision theory indirectly without me having to directly work them out. My suspicion currently focuses on communication theory, knowledge representation and data modelling (ontology). Rather than 'getting values out' I think values are most likely somehow implicitly built into the structure of ontology itself. |
| Re: What are the consequences of UD+ASSA? |
George Levy |
31/10/07 17:11 |
Could we relate the expansion of the universe to the decrease in measure of a given observer? High measure corresponds to a small universe and conversely, low measure to a large one. For the observer the decrease in his measure would be caused by all the possible mode of decay of all the nuclear particles necessary for his consciousness. Corresponding to this decrease, the radius of the observable universe increases to make the universe less likely. This would provide an experimental way to measure absolute measure. I am not a proponent of ASSA, rather I believe in RSSA and in a cosmological principle for measure: that measure is independent of when or where the observer makes an observation. However, I thought that tying cosmic expansion to measure may be an interesting avenue of inquiry. George Levy Rolf Nelson wrote:
>(Warning: This post assumes an familiarity with UD+ASSA and with the >cosmological Measure Problem.) > >Observational Consequences: > >1. Provides a possible explanation for the "Measure Problem" of why we >shouldn't be "extremely surprised" to find we live in a lawful >universe, rather than an extremely chaotic universe, or a homogeneous >cloud of gas. > >2. May help solve the Doomsday Argument in a finite universe, since >you probably have at least a little more "measure" than a typical >specific individual in the middle of a Galactic Empire, since you are >"easier to find" with a small search algorithm than someone surrounded >by enormous numbers of people. > >3. For similar reasons, may help solve a variant of the Doomsday >Argument where the universe is infinite. This variant DA asks, "if >there's currently a Galactic Empire 10000 Hubble Volumes away with an >immensely large number of people, why wasn't I born there instead of >here?" > >4. May help solve the Simulation Argument, again because a search >algorithm to find a particular simulation among all the adjacent >computations in a Galactic Empire is longer (and therefore, by UD >+ASSA, has less measure) than a search algorithm to find you. > >5. In basic UD+ASSA (on a typical Turing Machine), there is a probably >a strict linear ordering corresponding to when the events at each >point in spacetime were calculated; I would argue that we should >expect to see evidence of this in our observations if basic UD+ASSA is >true. However, we do not see any total ordering in the physical >Universe; quite the reverse: we see a homogeneous, isotropic Universe. >This is evidence (but not proof) that either UD+ASSA is completely >wrong, or that if UD+ASSA is true, then it's run on something other >than a typical linear Turing Machine. (However, if you still want use >a different machine to solve the "Measure Problem", then feel free, >but you first need to show that your non-Turing-machine variant still >solves the "Measure Problem.") > > >Decision Theory Consequences (Including Moral Consequences): > >Every decision algorithm that I've ever seen is prey to paradoxes >where the decision theory either crashes (fails to produce a >decision), or requires an agent to do things that are bizarre, self- >destructive, and evil. (If you like, substitute 'counter-intuitive' >for 'bizarre, self-destructive, and evil.') For example: UD+ASSA, >"Accepting the Simulation Argument", Utilitarianism without >discounting, and Utilitarianism with time and space discounting all >have places where they seem to fail. > >UD+ASSA, like the Simulation Argument, has the following additional >problem: while some forms of Utilitarianism may only fail in >hypothetical future situations (by which point maybe we'll have come >up with a better theory), UD+ASSA seems to fail *right here and now*. >That is, UD+ASSA, like the Simulation Argument, seems to call on you >to do bizarre, self-destructive, and evil things today. An example >that Yudowsky gave: you might spend resources on constructing a unique >arrow pointing at yourself, in order to increase your measure by >making it easier for a search algorithm to find you. > >Of course, I could solve the problem by deciding that I'd rather be >self-destructive and evil than be inconsistent; then I could consider >adopting UD+ASSA as a philosophy. But I think I'll pass on that >option. :-) > >So, more work would have to be done the morality of UD+ASSA before any >variant of UD+ASSA can becomes a realistically palatable part of a >moral philosophy. > >-Rolf > > >> > > > > |
| OM measure and universe size |
Russell Standish |
31/10/07 17:35 |
On Wed, Oct 31, 2007 at 05:11:01PM -0700, George Levy wrote: > > Could we relate the expansion of the universe to the decrease in > measure of a given observer? High measure corresponds to a small > universe and conversely, low measure to a large one. For the observer > the decrease in his measure would be caused by all the possible mode of > decay of all the nuclear particles necessary for his consciousness. > Corresponding to this decrease, the radius of the observable universe > increases to make the universe less likely. > > This would provide an experimental way to measure absolute measure. > > I am not a proponent of ASSA, rather I believe in RSSA and in a > cosmological principle for measure: that measure is independent of when > or where the observer makes an observation. However, I thought that > tying cosmic expansion to measure may be an interesting avenue of inquiry. > > George Levy > There is a relationship, though perhaps not quite what you think. The measure of an OM will be 2^{-C_O}, where C_O is the amount of information about the universe you know at that point in time (measured in bits). The physical complexity C of the universe at a point in time is in some sense the limit of all that is possible to know about the universe, ie C_O <= C. C is related to the size of the universe by the equation H = C + S, where S is the entropy of the universe (measured in bits), and H is the maximum possible entropy that would pertain if the universe were in equilibrium. H is a monotonically increasing function of the size of the universe - something like propertional to the volume (or similar - I forget the details). S is also an increasing function (due to the second law), but doesn't increase as fast as H. Consequently C increases as a function of universe age, and so C_O can be larger now than earlier in the universe, implying smaller OM measures. However, it remains to be seen whether the anthropic reasons for experiencing a universe 10^9 years and of large complexity we currently see is necessary... -- ---------------------------------------------------------------------------- A/Prof Russell Standish Phone 0425 253119 (mobile) Mathematics UNSW SYDNEY 2052 hpc...@hpcoders.com.au Australia http://www.hpcoders.com.au ----------------------------------------------------------------------------
|
| Re: What are the consequences of UD+ASSA? |
JohnM |
02/11/07 05:32 |
George, great. - Absolute measures? So you want to supersede the Archimede-Einstein wisdom ('gimme a fixed point"...to: total relativity) - which is OK with me. I like the way you approach questions (big deal for you<G>).
Main topic: Reverse Hubble? do we go towards a ;Big Bang', which is indeed a slow fade-out into a zero-point? (a slow No-Bang, indeed). I had questions about that expansionary idea, ingenious as it was. Brent did not like my skepticism, but I am no physicist and can take a physicist-put down.
I was missing the 'objective' (forgive me for this adjective) - all encompassing study to "exclude" ALL other possibilities for a redshift. (a topical impossibility). I had two little questions (never got answers):
1. do the 'atomic measures' (hypothetical as they may be) like distance "between" nucleus and electrons (calculational fairytale) also expand? or 2. does the physical story of today's intrinsic measures stay put and only the biggies expand?
In the first case nothing really happens, we just believe in a narrative.
So as much as I applaud your shrinking idea, it is still part of the narrative.
But it is a great idea. Thanks.
John M |
| Re: OM measure and universe size |
George Levy |
02/11/07 12:20 |
Russel,
We are trying to related the expansion of the universe to decreasing
measure. You have presented the interesting equation:
H = C + S
Let's try to assign some numbers.
1) Recently an article
appeared in New Scientist stating that we may be living "inside" a
black hole, with the event horizon being located at the limit of what
we can observe ie the radius of the current observable universe.
2) Stephen
Hawking showed that the entropy of a black hole is proportional to
its surface area.

where where k is Boltzmann's constant, and  is the Planck
length.
Thus we can say that a change in the Universe's radius corresponds to a
change in entropy dS. Therefore, dS/dt is proportional to dA/dt and to
8PR(dR/dt) R being the radius of the Universe and P = Pi. Let's assume
that dR/dt = c
Therefore
dS/dt = (k/4 L^2) 8PRc = 2kPRc/ L^2
Since Hubble
constant is 71 ± 4 ( km/ s)/ Mpc
which gives a size of the
Universe from the Earth to the edge of the visible universe. Thus R
= 46.5 billion light-years in any direction; this is the
comoving radius
of the visible universe. (Not the same as the age of the Universe
because of Relativity considerations)
Now I have trouble relating these facts to your equation H = C + S or
maybe to the differential version dH = dC + dS. What do you think? Can
we push this further?
George
|
| Re: OM measure and universe size |
Russell Standish |
02/11/07 23:56 |
On Fri, Nov 02, 2007 at 12:20:35PM -0700, George Levy wrote: > Russel, > > We are trying to related the expansion of the universe to decreasing > measure. You have presented the interesting equation: > > H = C + S > > Let's try to assign some numbers. > 1) Recently an article
> < http://space.newscientist.com/article/dn12853-black-holes-may-harbour-their-own-universes.html> > appeared in New Scientist stating that we may be living "inside" a black > hole, with the event horizon being located at the limit of what we can > observe ie the radius of the current observable universe. > 2) Stephen Hawking
> < http://en.wikipedia.org/wiki/Black_hole_thermodynamics> showed that the > entropy of a black hole is proportional to its surface area. >
> S_{BH} = \frac{kA}{4l_{\mathrm{P}}^2} > > where where k is Boltzmann's constant
> < http://en.wikipedia.org/wiki/Boltzmann%27s_constant>, and > l_{\mathrm{P}}=\sqrt{G\hbar / c^3} is the Planck length
> < http://en.wikipedia.org/wiki/Planck_length>. > > Thus we can say that a change in the Universe's radius corresponds to a > change in entropy dS. Therefore, dS/dt is proportional to dA/dt and to > 8PR(dR/dt) R being the radius of the Universe and P = Pi. Let's assume > that dR/dt = c > Therefore > > dS/dt = (k/4 L^2) 8PRc = 2kPRc/ L^2 >
> Since Hubble constant < http://en.wikipedia.org/wiki/Hubble%27s_law> is > 71 ą 4 (km < http://en.wikipedia.org/wiki/Kilometer>/s > < http://en.wikipedia.org/wiki/Second>)/Mpc > < http://en.wikipedia.org/wiki/Megaparsec> > > which gives a size of the Universe
> < http://en.wikipedia.org/wiki/Observable_universe> from the Earth to the > edge of the visible universe. Thus R = 46.5 billion light-years in any > direction; this is the comoving radius
> < http://en.wikipedia.org/wiki/Radius> of the visible universe. (Not the > same as the age of the Universe because of Relativity considerations) > > Now I have trouble relating these facts to your equation H = C + S or > maybe to the differential version dH = dC + dS. What do you think? Can > we push this further? > > George > I think that the formula you have above for S_{BH} is the value that should be taken for the H above. It is the maximum value that entropy can take for a volume the size of the universe. The internal observed entropy S, will of course, be much lower. I don't have a formula for it off-hand, but it probably involves the microwave background temperature. Cheers --
---------------------------------------------------------------------------- A/Prof Russell Standish Phone 0425 253119 (mobile) Mathematics UNSW SYDNEY 2052 hpc...@hpcoders.com.au Australia http://www.hpcoders.com.au ----------------------------------------------------------------------------
|
| Re: OM measure and universe size |
George Levy |
05/11/07 15:34 |
Sorry the nice equation formats did not make it past the server. Anyone
interested in the equations can find them at the associated wiki links.
George
Russell Standish wrote:
is
71 ± 4 (km <http://en.wikipedia.org/wiki/Kilometer>/s
<http://en.wikipedia.org/wiki/Second>)/Mpc
<http://en.wikipedia.org/wiki/Megaparsec>
which gives a size of the Universe
<http://en.wikipedia.org/wiki/Observable_universe> from the Earth to the
edge of the visible universe. Thus R = 46.5 billion light-years in any
direction; this is the comoving radius
<http://en.wikipedia.org/wiki/Radius> of the visible universe. (Not the
same as the age of the Universe because of Relativity considerations)
Now I have trouble relating these facts to your equation H = C + S or
maybe to the differential version dH = dC + dS. What do you think? Can
we push this further?
George
I think that the formula you have above for S_{BH} is the value that
should be taken for the H above. It is the maximum value that entropy
can take for a volume the size of the universe.
The internal observed entropy S, will of course, be much lower. I
don't have a formula for it off-hand, but it probably involves the
microwave background temperature.
Cheers
|
| Re: OM measure and universe size |
Russell Standish |
05/11/07 15:44 |
'twas perfectly readable to me, since it was bog-standard LaTeX notation which is a defacto standard for mathematical notation in email. Until someone figures out a way of getting all email clients to read and write mathML (which will probably be never), this is as good as it gets. Cheers -- ---------------------------------------------------------------------------- A/Prof Russell Standish Phone 0425 253119 (mobile) Mathematics UNSW SYDNEY 2052 hpc...@hpcoders.com.au Australia http://www.hpcoders.com.au ----------------------------------------------------------------------------
|
| Rép : Observer Moment = Sigma1-Sentences |
Bruno Marchal |
06/11/07 06:37 |
Hi David, I have almost finished the posts on the lobian machine I have promised. I have to make minor changes and to look a bit the spelling. I cannot do that this week, so I will send it next week. Thanks for your patience. I give you the plan, though, which I will actually also follow for the beginning (and the end) of the ULB-saturday course this year: 1) Cantor's diagonal 2) Does the universal digital machine exist? 3) Lobian machines, who and what are they? 4) The 1-person and the 3- machine. 5) Lobian machines' theology 6) Lobian machines' physics 7) Lobian machines' ethics
BTW, if some people are near Belgium, I have been invited for doing a talk on the UDA at a colloquium on "Logic and Reality" at Namur/Louvain in BELGIUM. The other talks seems quite interesting (too, if I may say :). Most will be done in english. Program and informations can be found here: http://www.logic-center.be/acts/logrea.html Best regards to David, and all of you Bruno http://iridia.ulb.ac.be/~marchal/
|
| Re: Rép : Observer Moment = Sigma1-Sentences |
David Nyman |
08/11/07 08:31 |
On Nov 6, 2:37 pm, Bruno Marchal <marc...@ulb.ac.be> wrote: > I have almost finished the posts on the lobian machine I have promised. > I have to make minor changes and to look a bit the spelling. I cannot > do that this week, so I will send it next week. Thanks for your > patience. Thanks - I'll keep an eye out. David |
| Re: Rép : Observer Moment = Sigma1-Sentences |
JohnM |
11/11/07 14:33 |
Bruno, I hope it will be accessible to me, too, by simple computerese software. John |
| Re: Rép : Observer Moment = Sigma1-Sentences |
Bruno Marchal |
13/11/07 06:52 |
Hi John, Le 11-nov.-07, à 23:33, John Mikes a écrit :
> Bruno, I hope it will be accessible to me, too, by simple computerese > software. Normally there should be no difficulties. My goal is not to explain all the technics, but the minimal things which I estimate to be necessary for having a basic general idea of what is going on.
My first goal, perhaps my main goal, is to explain Church Thesis CT. To explain why CT is a very strong hypothesis, with a uniform deep impact on everything, and mainly on "theories of everything". I want also to explain more clearly the difference between Tegmark, Schmidhuber, and "comp", etc. But this needs a minimal amount of "modern math", so as to make clear Cantor's role, and then Church, Kleene. Not really the time today, but hopefully (normally) I will have more time tomorrow, Thanks for letting me know your interest, and your patience, Best, Bruno http://iridia.ulb.ac.be/~marchal/
|
| Bijections (was OM = SIGMA1) |
Bruno Marchal |
14/11/07 07:21 |
Hi David, Hi John, OK, here is a first try. Let me know if this is too easy, too difficult, or something in between. The path is not so long, so it is useful to take time on the very beginning. I end up with some exercice. I will give the solutions, but please try to be aware if you can or cannot do them, so as not missing the train. John, if you have never done what has been called "modern math", you could have slight notation problem, please ask any question. I guess for some other the first posts in this thread could look too much simple. Be careful when we will go from this to the computability matter. I recall the plan, where I have added the bijection thread: Plan 0) Bijections 1) Cantor's diagonal 2) Does the universal digital machine exist?
And for much later, if people are interested or ask question: 3) Lobian machines, who and/or what are they? 4) The 1-person and the 3- machine. 5) Lobian machines' theology 6) Lobian machines' physics 7) Lobian machines' ethics
But my main goal first is to explain that Church thesis is a very strong postulate. I need first to be sure you have no trouble with the notion of bijection. ================ 0) Bijections Suppose you have a flock of sheep. Your neighbor too. You want to know if you have more, less or the same number of sheep, but the trouble is that neither you nor your neighbor can count (nor anyone around). Amazingly enough, perhaps, it is still possible to answer that question, at least if you have enough pieces of rope. The idea consists in attaching one extremity of a rope to one of your sheep and the other extremity to one of the neighbor's sheep, and then to continue. You are not allow to attach two ropes to one sheep, never. In the case you and your neighbor have a different number of sheep, some sheep will lack a corresponding sheep at the extremity of their rope, so that their ropes will not be attached to some other sheep. Exemple (your flock of sheep = {s, r, t, u}, and the sheep of the neighborgh = {a, b, c, d, e, f}. s --------- a r ---------- d u --------- c t ----------- f ------------e ------------b and we see that the neighbor has more sheep than you, because b and e have their ropes unable to be attached to any remaining sheep you have, and this because there are no more sheep left in your flock. You have definitely less sheep. In case all ropes attached in that way have a sheep well attached at both extremities, we can say that your flock and your neighbor's flock have the same number of elements, or same cardinality. In that case, the ropes represent a so called one-one function, or bijection, between the two flocks. If you have less sheep than your neighbor, then there is a bijection between your flock and a subset of your neighbor's flock. If those flocks constitute a finite set, the existence of a bijection between the two flocks means that both flocks have the same number of sheep, and this is the idea that Cantor will generalize to get a notion of "same number of element" or "same cardinality" for couples of infinite sets. Given that it is not clear, indeed, if we can count the number of element of an infinite set, Cantor will have the idea of generalizing the notion of "same number of elements", or "same cardinality" by the existence of such one-one function. The term "bijection" denotes "one-one function". Definition: A and B have same cardinality (size, number of elements) when there is a bijection from A to B.
Now, at first sight, we could think that all *infinite* sets have the same cardinality, indeed the "cardinality" of the infinite set N. By N, I mean of course the set {0, 1, 2, 3, 4, ...} By E, I mean the set of even number {0, 2, 4, 6, 8, ...} Galileo is the first, to my knowledge to realize that N and E have the "same number of elements", in Cantor's sense. By this I mean that Galileo realized that there is a bijection between N and E. For example, the function which sends x on 2*x, for each x in N is such a bijection. Now, instead of taking this at face value like Cantor, Galileo will instead take this as a warning against the use of the infinite in math or calculus. Confronted to more complex analytical problems of convergence of Fourier series, Cantor knew that throwing away infinite sets was too pricy, and on the contrary, will consider such problems as a motivation for its "set theory". Dedekind will even define an infinite set by a set which is such that there is a bijection between itself and some proper subset of himself. By Z, I mean the set of integers {..., -3, -2, -1, 0, 1, 2, 3, ...} Again there is a bijection between N and Z. For example, 0 1 2 3 4 5 6 7 8 9 10 .... 0 -1 1 -2 2 -3 3 -4 4 -5 5 ... or perhaps more clearly (especially if the mail does not respect the "blank" uniformely; the bijection, like all function, is better represented as set of couples: bijection from N to Z = {(0,0) (1, -1) (2, 1) (3 -2) (4, 2) (5, -3) (6, 3) ... }. Because everyone know the sequence 0, 1, 2, 3, 4, 5, ... we can also describe a bijection between N and Z (say) just by the sequence of images: 0 -1 1 -2 2 -3 3 -4 4 -5 5 ... That bijection can also be given by a rule: send even number x on x/2, and send odd numbers on x -((x+1)/2). But it is not necessary to get those rules to be convinced, the drawing is enough once you interpret it genuinely. By AXB, I mean the set of couples (x, y) with x in A and y in B. It is natural to put them in a cartesian plane. For exemple, if A = {0, 1} and B = {a, b, c}, then AXB = {(0, a) (1, a) (0, b) (1, b) (0, c) (1, c)}, and is best represented by
(0, c) (1, c) (0, b) (1, b) (0, a) (1, a) You see that if A is finite and has n elements and if B is finite and has m elements, then AXB is finite and has m*n elements. Yet, again, NXN "has the same number of elements" that N. NXN is obviously the infinite extension of the following 4X4 approximation:
... (0,3) (1,3) (2,3) (3,3)... (0,2) (1,2) (2,2) (3,2)... (0,1) (1,1) (2,1) (3,1)... (0,0) (1,0) (2,0) (3,0)... Do you see a bijection between N and NXN ? Here is one, which I will call the zigzagger (draw the picture above and draw the link between the couples, a bit like in little children drawing puzzles, so as to see the zigzag clearly). (0,0) (0,1) (1,0) (2,0) (1,1) (0,2) (0,3) (1,2) (2,1) (3,0) (4,0) (3,1) (2,2) (1,3) (0,4) ... Here is another one, due to Cantor, I think. To draw it you will have to raise the pen. (0,0) (0,1) (1,0) (0,2) (1,1) (2,0) (0,3) (1,2) (2,1) (3,0) (0,4) (1,3) (2,2) (3,1) (4,0) ... The inverse of that bijection, which exists and is of course a bijection from NXN to N has a nice quasi polynomial presentation. (x, y) is send on the half of (x+y)^2 + 3x + y. You see that (4,0) is send to 14 (ok ?, it is not 13 because we start from zero), and indeed the half of (4+0)^2 + 3*4 +0 is 14. And ZXZ extends in a similar way : ... ... (-3, 3) (-2, 3) (-1,3) (0,3) (1,3) (2,3) (3,3)... ... (-3, 2) (-2, 2) (-1,2) (0,2) (1,2) (2,2) (3,2)... ... (-3, 1) (-2, 1) (-1,1) (0,1) (1,1) (2,1) (3,1)... ... (-3,0) (-2, 0) (-1,0) (0,0) (1,0) (2,0) (3,0)... ... (-3,-1) (-2,-1)(-1,-1) (0,-1) (1,-1) (2,-1) (3,-1)... ... (-3,-2) (-2,-2)(-1,-2) (0,-2) (1,-2) (2,-1) (3,-2)... ... (-3,-3) (-2,-3)(-1,-3) (0,-3) (1,-3) (2,-1) (3,-3)... ... Do you see a bijection between N and ZXZ ? Here is the spiral-bijection, or spiraler: (0,0) (0,1) (-1,1) (-1,0) (-1,-1) (0,-1) (1,-1) (1,0) (1,1) (1,2) (0,2) (-1,2) (-2,2) (-2,-1) .... You see: a bijection between N and ZXZ assigns to each natural number one and only one couple of integers, and this in such a way that we are sure that all couples of integers is the image of a natural number by that bijection. Little exercises:
1) is there a bijection between N and N? 2) Q is the set of rational numbers, that is length of segment which can be measured by the ratio of natural numbers (or integers). (equivalently: = the real with repetitive decimals, like 0,99999999...., or 345,78123123123123123...). By the way, could find n and m (in N) such that n/m = 345,78123123123123123... ? (and could you explain why 1,00000000.... = 0, 99999999....?). But this will not been used later. Can you see that there is a bijection between N and Q? Hint: transform the bijection between N and NXN into a bijection between N and Q. 3) The disjoint union of two sets A and B = their formal union together with a relabelling so as to distinguish the elements: exemple: N disjoint-union-with N = {0, 1, 2, 3, ...} U {0', 1', 2', 3', ...}, which I will write N U N' (read: N union N prime). Is there a bijection between N and (with obvious notation) N U N' U N'' U N''' U ...? I will write NXNXN for NX(NXN) can you build a bijection between N and NXNXN ? can you build a bijection between N and NXNXNXN ? can you build a bijection between N and NXNXNXNXN ? 4) Very important for the sequel. Let A be a finite alphabet (for exemple: A = {0,1}). Let us denote by A° the set of all finite words build on A. By a (finite) word a mean any finite sequence of elements taken in the alphabet (like 000, or 10101, or 1101000011010110). Is there a bijection between N and A°, with A = {0,1}. Solutions will be given later, but ask for question if there are any problem. Don't hesitate to tell me the mistakes, which always exist! You can also ask question about motivation, like, "for God sake why are you explaining us all this". Try also to keep those posts for later considerations. ================ http://iridia.ulb.ac.be/~marchal/
|
| Re: Bijections (was OM = SIGMA1) |
Torgny Tholerus |
14/11/07 08:23 |
Bruno Marchal skrev: > 0) Bijections > > Definition: A and B have same cardinality (size, number of elements) > when there is a bijection from A to B. > > Now, at first sight, we could think that all *infinite* sets have the > same cardinality, indeed the "cardinality" of the infinite set N. By N, > I mean of course the set {0, 1, 2, 3, 4, ...} >
What do you mean by "..."? > By E, I mean the set of even number {0, 2, 4, 6, 8, ...} > > Galileo is the first, to my knowledge to realize that N and E have the > "same number of elements", in Cantor's sense. By this I mean that > Galileo realized that there is a bijection between N and E. For > example, the function which sends x on 2*x, for each x in N is such a > bijection. >
What do you mean by "each x" here? How do you prove that each x in N has a corresponding number 2*x in E? If m is the biggest number in N, then there will be no corresponding number 2*m in E, because 2*m is not a number. > Now, instead of taking this at face value like Cantor, Galileo will > instead take this as a warning against the use of the infinite in math > or calculus. >
-- Torgny Tholerus |
| Re: Bijections (was OM = SIGMA1) |
Bruno Marchal |
15/11/07 03:31 |
Le 14-nov.-07, à 17:23, Torgny Tholerus a écrit : > > Bruno Marchal skrev: >> 0) Bijections >> >> Definition: A and B have same cardinality (size, number of elements) >> when there is a bijection from A to B. >> >> Now, at first sight, we could think that all *infinite* sets have the >> same cardinality, indeed the "cardinality" of the infinite set N. By >> N, >> I mean of course the set {0, 1, 2, 3, 4, ...} >> > What do you mean by "..."? Are you asking this as a student who does not understand the math, or as a philospher who, like an ultrafinist, does not believe in the potential infinite (accepted by mechanist, finistist, intuitionist, etc.).
I have already explained that the meaning of "...'" in {I, II, III, IIII, IIIII, IIIIII, IIIIIII, IIIIIIII, IIIIIIIII, ...} is *the* mystery. A beautiful thing, which is premature at this stage of the thread, is that accepting the usual meaning of "..." , then we can mathematically explained why the meaning of "..." has to be a mystery. >> By E, I mean the set of even number {0, 2, 4, 6, 8, ...} >> >> Galileo is the first, to my knowledge to realize that N and E have the >> "same number of elements", in Cantor's sense. By this I mean that >> Galileo realized that there is a bijection between N and E. For >> example, the function which sends x on 2*x, for each x in N is such a >> bijection. >> > What do you mean by "each x" here?
I mean "for each natural number".
> > How do you prove that each x in N has a corresponding number 2*x in E? > If m is the biggest number in N,
There is no biggest number in N. By definition of N we accept that if x is in N, then x+1 is also in N, and is different from x.
> then there will be no corresponding > number 2*m in E, because 2*m is not a number. Of course, but you are not using the usual notion of numbers. If you believe that the usual notion of numbers is wrong, I am sorry I cannot help you.
Bruno >> Now, instead of taking this at face value like Cantor, Galileo will >> instead take this as a warning against the use of the infinite in math >> or calculus. >> > -- > Torgny Tholerus > > > >
http://iridia.ulb.ac.be/~marchal/ |
| Re: Bijections (was OM = SIGMA1) |
Torgny Tholerus |
15/11/07 05:45 |
Bruno Marchal skrev:
Le 14-nov.-07, à 17:23, Torgny Tholerus a écrit :
What do you mean by "..."?
Are you asking this as a student who does not understand the math, or
as a philospher who, like an ultrafinist, does not believe in the
potential infinite (accepted by mechanist, finistist, intuitionist,
etc.).
I am asking as an ultrafinitist.
I have already explained that the meaning of "...'" in {I, II, III,
IIII, IIIII, IIIIII, IIIIIII, IIIIIIII, IIIIIIIII, ...} is *the*
mystery.
Do you have the big-black-cloud interpretation of "..."? By that I
mean that there is a big black cloud at the end of the visible part of
universe, and the sequence of numbers is disappearing into the cloud,
so that you can only see the numbers before the cloud, but you can not
see what happens at the end of the sequence, because it is hidden by
the cloud.
For
example, the function which sends x on 2*x, for each x in N is such a
bijection.
What do you mean by "each x" here?
I mean "for each natural number".
What do you mean by "each" in the sentence "for each natural number"?
How do you define ALL natural numbers?
How do you prove that each x in N has a corresponding number 2*x in E?
If m is the biggest number in N,
There is no biggest number in N. By definition of N we accept that if x
is in N, then x+1 is also in N, and is different from x.
How do you know that m+1 is also in N? You say that for ALL x then x+1
is included in N, but how do you prove that m is included in "ALL x"?
If you say that m is included in "ALL x", then you are doing an illegal
deduction, and when you do an illegal deduction, then you can prove
anything. (This is the same illegal deduction that is made in the
Russell paradox.)
then there will be no corresponding
number 2*m in E, because 2*m is not a number.
Of course, but you are not using the usual notion of numbers. If you
believe that the usual notion of numbers is wrong, I am sorry I cannot
help you.
I am using the usual notion of numbers. But m+1 is not a number. But
you can define a new concept: "number-2", such that m+1 is included in
that new concept. And you can define a new set N2, that contains all
natural numbers-2. This new set N2 is bigger than the old set N, that
only contains all natural numbers.
--
Torgny Tholerus
|
| Re: Bijections (was OM = SIGMA1) |
Quentin Anciaux |
15/11/07 11:38 |
Hi, Le Thursday 15 November 2007 14:45:24 Torgny Tholerus, vous avez écrit : >> Bruno Marchal skrev: > Le 14-nov.-07, à 17:23, Torgny Tholerus a écrit :
>>> What do you mean by "each x" here? > > > > >I mean "for each natural number". > > > What do you mean by "each" in the sentence "for each natural number"? How > do you define ALL natural numbers? > There is a natural number 0. Every natural number a has a natural number successor, denoted by S(a). There is no natural number whose successor is 0. Distinct natural numbers have distinct successors: if a ≠ b, then S(a) ≠ S(b). You need at least the successor axiom. N = {0 ,1 ,2 ,3 ,... ,N ,N+1, ..} All natural numbers are defined by the above. > > > How do you prove that each x in N has a corresponding number 2*x in E? > If m is the biggest number in N, By definition there exists no biggest number unless you add an axiom saying there is one but the newly defined set is not N. Quentin Anciaux -- All those moments will be lost in time, like tears in the rain.
|
| Re: Bijections (was OM = SIGMA1) |
Torgny Tholerus |
16/11/07 00:33 |
Quentin Anciaux skrev:
Hi,
Le Thursday 15 November 2007 14:45:24 Torgny Tholerus, vous avez écrit :
What do you mean by "each" in the sentence "for each natural number"? How
do you define ALL natural numbers?
There is a natural number 0.
Every natural number a has a natural number successor, denoted by S(a).
What do you mean by "Every" here? Can you give a *non-circular*
definition of this word? Such that: "By every natural number I mean
{1,2,3}" or "By every naturla number I mean every number between 1 and
1000000". (This last definition is non-circular because here you can
replace "every number" by explicit counting.)
How do you prove that each x in N has a corresponding number 2*x in E?
If m is the biggest number in N,
By definition there exists no biggest number unless you add an axiom saying
there is one but the newly defined set is not N.
I can prove by induction that there exists a biggest number:
A) In the set {m} with one element, there exists a biggest number, this
is the number m.
B) If you have a set M of numbers, and that set have a biggest number
m, and you add a number m2 to this set, then this new set M2 will have
a biggest number, either m if m is bigger than m2, or m2 if m2 is
bigger than m.
C) The induction axiom then says that every set of numbers have a
biggest number.
Q.E.D.
--
Torgny Tholerus
|
| Re: Bijections (was OM = SIGMA1) |
Quentin Anciaux |
16/11/07 01:30 |
Le Friday 16 November 2007 09:33:38 Torgny Tholerus, vous avez écrit : > Quentin Anciaux skrev: > Hi, > > Le Thursday 15 November 2007 14:45:24 Torgny Tholerus, vous avez écrit : > > What do you mean by "each" in the sentence "for each natural number"? > How do you define ALL natural numbers? > > > > There is a natural number 0. > Every natural number a has a natural number successor, denoted by S(a). > > > What do you mean by "Every" here? Can you give a *non-circular* > definition of this word? Such that: "By every natural number I mean > {1,2,3}" or "By every naturla number I mean every number between 1 and > 1000000". (This last definition is non-circular because here you can > replace "every number" by explicit counting.) I do not see circularity here... every means every, it means all natural numbers possess this properties ie (having a successor), that means by induction that N does contains an infinite number of elements, if it wasn't the case that would mean that there exists a natural number which doesn't have a successor... well as we have put explicitly the successor rule to defined N I can't see how to change that without changing the axioms. > > > How do you prove that each x in N has a corresponding number 2*x in E? > If m is the biggest number in N, > > > By definition there exists no biggest number unless you add an axiom saying > there is one but the newly defined set is not N. > > > I can prove by induction that there exists a biggest number: > > A) In the set {m} with one element, there exists a biggest number, this is > the number m. B) If you have a set M of numbers, and that set have a > biggest number m, and you add a number m2 to this set, then this new set M2 > will have a biggest number, either m if m is bigger than m2, or m2 if m2 is > bigger than m. C) The induction axiom then says that every set of numbers > have a biggest number. > > Q.E.D. > > -- > Torgny Tholerus Hmm I don't understand... This could only work on finite set of elements. I don't see this as a proof that N is finite (because it *can't* be by *definition*). Quentin Anciaux -- All those moments will be lost in time, like tears in the rain.
|
| Re: Bijections (was OM = SIGMA1) |
Bruno Marchal |
16/11/07 03:50 |
Le 15-nov.-07, à 14:45, Torgny Tholerus a écrit : > Bruno Marchal skrev:Le 14-nov.-07, à 17:23, Torgny Tholerus a écrit : >> >> >>> What do you mean by "..."? >>> >> >> Are you asking this as a student who does not understand the math, or >> as a philospher who, like an ultrafinist, does not believe in the >> potential infinite (accepted by mechanist, finistist, intuitionist, >> etc.). >> > > I am asking as an ultrafinitist.
Fair enough. I am not sure there are many ultrafinitists on the list, but just to let John Mikes and Norman to digest the bijection post, I will say a bit more. A preliminary remark is that I am not sure an ultrafinitist can really assert he is ultrafinitist without acknowledging that he does have a way to give some meaning on "...". But I have a more serious question below.
> >> I have already explained that the meaning of "...'" in {I, II, III, >> IIII, IIIII, IIIIII, IIIIIII, IIIIIIII, IIIIIIIII, ...} is *the* >> mystery. >> > > Do you have the big-black-cloud interpretation of "..."? By that I > mean that there is a big black cloud at the end of the visible part of > universe,
Concerning what I am trying to convey, this is problematic. The word "universe" is problematic. The word "visible" is also problematic.
> and the sequence of numbers is disappearing into the cloud, so that > you can only see the numbers before the cloud, but you can not see > what happens at the end of the sequence, because it is hidden by the > cloud. I don't think that math is about seeing. I have never seen a number. It is a category mistake. I can interpret sometimes some symbol as refering to number, but that's all.
> >> >> >>>> For >>>> example, the function which sends x on 2*x, for each x in N is such >>>> a >>>> bijection. >>>> >>>> >>> What do you mean by "each x" here? >>> >> >> I mean "for each natural number". >> > > What do you mean by "each" in the sentence "for each natural > number"? How do you define ALL natural numbers?
By relying on your intuition of "finiteness". I take 0 as denoting a natural number which is not a successor. I take s(0) to denote the successor of 0. I accept that any number obtained by a *finite* application of the successor operation is a number. I accept that s is a bijection from N to N \ {0}, and things like that.
> >> >> >>> How do you prove that each x in N has a corresponding number 2*x in >>> E? >>> If m is the biggest number in N, >>> >> >> There is no biggest number in N. By definition of N we accept that if >> x >> is in N, then x+1 is also in N, and is different from x. >> > > How do you know that m+1 is also in N?
By definition. > You say that for ALL x then x+1 is included in N, but how do you prove > that m is included in "ALL x"?
I say "for all x" means "for all x in N". > > If you say that m is included in "ALL x", then you are doing an > illegal deduction, and when you do an illegal deduction, then you can > prove anything. (This is the same illegal deduction that is made in > the Russell paradox.)
? (if you believe this then you have to accept that Peano Arithmetic, or even Robinson arithmetic) is inconsistent. Show me the precise proof.
> >> >> >>> then there will be no corresponding >>> number 2*m in E, because 2*m is not a number. >>> >> >> Of course, but you are not using the usual notion of numbers. If you >> believe that the usual notion of numbers is wrong, I am sorry I cannot >> help you. >> > > I am using the usual notion of numbers. You are not. By definition of the usual natural numbers, all have a successor.
> But m+1 is not a number.
This means that you believe there is a finite sequence of "s" of the type
A = s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s( s(s(s(s( ....s(0)))))))))))))))))))))...) where "..." here represents a finite sequence, and which is such that s(A) is not a number. > But you can define a new concept: "number-2", such that m+1 is > included in that new concept. And you can define a new set N2, that > contains all natural numbers-2. This new set N2 is bigger than the > old set N, that only contains all natural numbers.
Torgny, have you followed my "fairy tale" which I have explain to Tom Caylor. There I have used transfinite sequence of growing functions to name a big but finite natural number, which I wrote F_superomega(999), or OMEGA+[OMEGA]+OMEGA. My "serious" question is the following: is your "biggest number" less, equal or bigger than a well defined finite number like F_superomega(999).
If yes, then a big part of the OM = SIGMA_1 thread will be accessible to you, except for the final conclusion. Indeed, you will end up with a unique finite bigger universal machine (which I doubt). If not, let us just say that your ultrafinitist hypothesis is too strong to make it coherent with the computationalist hypo. It means that you have a theory which is just different from what I propose. And then I will ask you to be "ultra-patient", for I prefer to continue my explanation, and to come back on the discussion on hypotheses after. OK. Actually, my conversation with Tom was interrupted by Norman who fears people leaving the list when matter get too much technical; but I was about to introduce a second fairy, capable of non constructive reasoning. I recall that the first fairy asked for a big constructively nameable number. The second fairy was supposed to ask the same except that she drops the constructivity requirement. In that case you can use even higher infinities to name (non constructively) a natural number. This gives a way to name finite number still *vastly* bigger than F_superomega(999). In case your bigger natural number is bigger than such monstruosity, then the whole of the thread will be understandable, although comp can be made explicitly wrong (in that case) but somehow approximable in practice (even in the local comp practice). Of course if your bigger natural number is little than 100^(100^(100^(100^(100^(100^(100^(100^100)], which is already far beyond any empirical numbers motivated by the observable empirical world, then we can only conclude that we have very different theory. BTW, do you agree that 100^(100^(100^(100^(100^(100^(100^(100^100)], and 100^(100^(100^(100^(100^(100^(100^(100^100)] +1 are numbers? I am just curious, Bruno
http://iridia.ulb.ac.be/~marchal/
|
| Re: Bijections (was OM = SIGMA1) |
Bruno Marchal |
16/11/07 06:46 |
Le 16-nov.-07, à 09:33, Torgny Tholerus a écrit : >>>> >> There is a natural number 0. >> Every natural number a has a natural number successor, denoted by >> S(a). >> > > What do you mean by "Every" here? > Can you give a *non-circular* definition of this word? Such that: "By > every natural number I mean {1,2,3}" or "By every naturla number I > mean every number between 1 and 1000000". (This last definition is > non-circular because here you can replace "every number" by explicit > counting.) > >> >>> How do you prove that each x in N has a corresponding number 2*x in >>> E? >>> If m is the biggest number in N, >>> >> By definition there exists no biggest number unless you add an axiom >> saying >> there is one but the newly defined set is not N. >> > > I can prove by induction that there exists a biggest number: > > A) In the set {m} with one element, there exists a biggest number, > this is the number m. > B) If you have a set M of numbers, and that set have a biggest number > m, and you add a number m2 to this set, then this new set M2 will have > a biggest number, either m if m is bigger than m2, or m2 if m2 is > bigger than m. > C) The induction axiom then says that every set of numbers have a > biggest number.
What do you mean by "every" here? You just give us a non ultrafinitistic proof that all numbers are finite, not that the set of all finite number is finite.
Bruno
> > Q.E.D. > > -- > Torgny Tholerus > > > > http://iridia.ulb.ac.be/~marchal/
|
| Re: Bijections (was OM = SIGMA1) |
Torgny Tholerus |
16/11/07 09:14 |
Bruno Marchal skrev:
Le 15-nov.-07, à 14:45, Torgny Tholerus a écrit :
But m+1 is not a number.
This means that you believe there is a finite sequence of "s" of the
type
A =
s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(
....s(0)))))))))))))))))))))...)
where "..." here represents a finite sequence, and which is such that
s(A) is not a number.
Yes, exactly. When you construct the set of ALL natural numbers N, you
have to define ALL these numbers. And you can only define a finite
number of numbers. See more explanations below.
BTW, do you agree that 100^(100^(100^(100^(100^(100^(100^(100^100)],
and 100^(100^(100^(100^(100^(100^(100^(100^100)] +1 are numbers? I am
just curious,
Yes, I agree. All explicitly given numbers are numbers. The biggest
number is bigger than all by human beeings explicitly given numbers.
If you define the set of all natural numbers N, then you can pull out
the biggest number m from that set. But this number m has a different
"type" than the ordinary numbers. (You see that I have some sort of
"type theory" for the numbers.) The ordinary deduction rules do not
hold for numbers of this new type. For all ordinary numbers you can
draw the conclusion that the successor of the number is included in N.
But for numbers of this new type, you can not draw this conclusion.
You can say that all ordinary natural numbers are of type 0. And the
biggest natural number m, and all numbers you construct from that
number, such that m+1, 2*m, m/2, and so on, are of type 1. And you can
construct a set N1 consisting of all numbers of type 1. In this set
there exists a biggest number. You can call it m1. But this new
number is a number of type 2.
There is some sort of "temporal" distinction between the numbers of
different type. You have to "first" have all numbers of type 0,
"before" you can construct the numbers of type 1. And you must have
all numbers of type 1 "before" you can construct any number of type 2,
and so on.
The construction of numbers of type 1 presupposes that the set of all
numbers of type 0 is fixed. When the set N of all numbers of type 0 is
fixed, then you can construct new numbers of type 1.
It may look like a contradiction to say that m is included in N, and to
say that all numbers in N have a successor in N, and to say that m have
no successor in N. But it is not a constrdiction because the rule "all
numbers in N have a successor in N" can be expanded to "all numbers of
type 0 in N have a successor in N". And because m is a number of type
1, then that rule is not applicable to m.
--
Torgny
|
| Re: Bijections (was OM = SIGMA1) |
Brent |
16/11/07 09:41 |
Bruno Marchal wrote: > ... > If not, let us just say that your ultrafinitist hypothesis is too > strong to make it coherent with the computationalist hypo. It means > that you have a theory which is just different from what I propose. > And then I will ask you to be "ultra-patient", for I prefer to > continue my explanation, and to come back on the discussion on > hypotheses after. OK. > > Actually, my conversation with Tom was interrupted by Norman who fears > people leaving the list when matter get too much technical;
Pay no attention to Norman. :-) I attend to this list because I learn things from it and I learn a lot from your technical presentations. I'm also doubtful of infinities, but they make things simpler; so my attitude is, let's see where the theory takes us. Brent Meeker |
| The big-black-cloud-interpretation. |
Torgny Tholerus |
16/11/07 09:43 |
Bruno Marchal skrev:
Le 15-nov.-07, à 14:45, Torgny Tholerus a écrit :
Do you have the big-black-cloud interpretation of "..."?
By that I
mean that there is a big black cloud at the end of the visible part of
universe,
Concerning what I am trying to convey, this is problematic. The word
"universe" is problematic. The word "visible" is also problematic.
and the sequence of numbers is disappearing into the
cloud,
so that you can only see the numbers before the cloud, but you can
not see what happens at the end of the sequence, because it is hidden
by the cloud.
I don't think that math is about seeing. I have never seen a number.
It is a category mistake. I can interpret sometimes some symbol as
refering to number, but that's all.
A way to prove the consistency of a theory is to make a "visualization"
of the theory. If you can visualize all that happens in the theory,
then you know the theory is consistent.
To visualize the natural numbers, you can think of them as a long
sequence {0,1,2,3,4,5,...}, and this sequence is going far, far, away.
But you can only visualize finite sequences. So you can think that you
have a finite sequence of numbers, and you have a big black cloud far,
far, away. You see the first part of the sequence {0,1,2,...,m} before
the cloud. But inside the cloud you can imagine that you have the
finite sequence {m+1,m+2,...,4*m-1,4*m}. This whole sequence
{0,1,2,...,m,m+1,...4*m} is what you call the set N of all natural
numbers.
From that set N you construct the true subset
{0,2,4,6,...,2*m,2*m+2,...,4*m}, which you call the set E of all even
numbers. The visible part of the set E is then {0,2,4,...,2*m}, and
the hidden part of that sequence is {2*m+2,...,4*m}.
Now you define a new concept INNFINITE, that is defined by:
If you have a bijection from all visible numbers of a set S, to all
visible numbers of a true subset of S, then you say that the set S in
INNFINITE.
Then you can use this concept INNFINITE, and you will get a consistent
theory with no contradictions, because you have a finite visualization
of this theory.
--
Torgny
|
| Re: Bijections (was OM = SIGMA1) |
Bruno Marchal |
19/11/07 06:58 |
Le 16-nov.-07, à 18:41, meekerdb (Brent Meeker) a écrit : Fair enough, thanks. I give the solution of the little exercises on the notion of bijection. 1) is there a bijection between N and N? Of course! The identity function is a bijection from N to N, and actually any identity function defined on any set is a bijection from that set with itself. Here is a drawing of a bijection from N to N (I don't represent the "ropes" ...) 0 1 2 3 4 5 6 7 8 ... 0 1 2 3 4 5 6 7 8 ... So all sets "have the same number of element" than itself. 2) Q is the set of rational numbers, that is length of segment which can be measured by the ratio of natural numbers (or integers).
There is a bijection between N and Q. We have already seen a bijection between N and ZXZ, which I called the spiraller (I put only the image of the bijection, imagine the rope 0----(0,0), 1-----(0,1), 2-----(-1,1), ... : (0,0) (0,1) (-1,1) (-1,0) (-1,-1) (0,-1) (1,-1) (1,0) (1,1) (1,2) (0,2) (-1,2) (-2,2) (-2,-1) .... We can transform that bijection between N and ZXZ into a bijection between N and Q in the following way. We start from the bijection above between N and ZXZ, but we interpret (x,y) as the fraction x/y, throwing it out in case we already met the corresponding rational number, or when we met an indeterminate fraction (like 0/0) or spurious one (like 1/0, 2/0, ...), this gives first 0/0 0/1 -1/1 -1/0 -1/-1 0/-1 1/-1 1/0 1/1 1/2 0/2 -1/2 -2/2 -2/1 ...., and after the throwing out of the repeated rationals: 0 -1 1 1/2 -1/2 -2 ... OK? 3) We have seen a bijection between N and NXN. We can use it to provide a bijection between N and NX(NXN). Indeed, you can zigzag on NX(NXN) like we have zigzag on NXN, starting from:
... 5 4 3 2 1 0 (0,0) (0,1) (1,0) (2,0) (1,1) (0,2) (0,3) (1,2) (2,1) (3,0) (4,0)
(3,1) ... All right? Obviously there is a bijection between NXNXN and NX(NXN)); just send (x,y,z) on (x,(y,z)). In the same manner, you can show the existence of a bijection and any of NXNXNXN, NXNXNXNXN, NXNXNXNXNXN, .... 4) (The one important for the sequel). Take any finite alphabet, like {0,1}, {a, b}, {a, b, c, ... z} or all keyboard keys. Then the set of all finite words build on any such alphabet is in bijection with N. Indeed, to be sure of enumerating all the words, on {a,b,c}, say, just enumerate the words having length 1, then length 2, etc. And order just alphabetically the words having the same length. On the alphabet {a,b,c}, this gives a, b, c, aa, ab, ac, ba, bb, bc, ca, cb, cc, aaa, aab, aac, aba, abb, abc, aca, acb, acc, baa, bab, bac, bba, bbb, bbc, bca, bcb, bcc, ... I hope you see that this gives a bijection between N and A° (the set of words on A, which in this example is {a, b, c}. Purist would add the empty word in front. From this you can give another proof that Q is enumerable (= in bijection with N). Indeed all rational numbers, being a (reduced) fraction, can be written univocally as a word in the finite alphabet {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, /}. Example 456765678 / 9898989 (I have added blank for reason of readibility, but those are not symbols taken from the alphabet). Another important example, the set of all programs in any programming language. For some language, like Python for example, you have to put explicitly the "enter", or "carriage return" key symbol in the alphabet, of course. 5) Let N be the set {0, 1, 2, 3, 4, 5, 6, ...}, as usual. Let N' be a sort of copy or relabeling of N, that is N' = {0', 1', 2', 3', 4', 5', 6', ...}. What are really object like 5' is not relevant except that we consider that any x is different from any x'. (usually mathematician formalize N' by NX{0}, or NX{1}, but that are details).
Is N U N' in bijection with N ? Sure, NUN' = {0, 0', 1, 1', 2, 2', 3, 3', 4, 4', 5, 5', 6, 6', 7, 7', 8, 8', ...}, and 0, 0', 1, 1', 2, 2', 3, 3', 4, 4', 5, 5', 6, 6', 7, 7', 8, 8', ... is indeed an enumeration of NUN' (bijection from N to the set NUN'). You can see it as the result of a zigzagging between 0, 1, 2, 3, 4, 5, 6, ... 0', 1', 2', 3', 4', 5', 6', ... (do the drawing if you want to see the zigzag). Is the infinite union NUN'UN''UN'''UN''''U ... still enumerable? Yes, just zigzag on
0, 1, 2, 3, 4, 5, 6, ... 0', 1', 2', 3', 4', 5', 6', .. 0'', 1'', 2'', 3'', 4'', 5'', 6'', ... 0''', 1''', 2''', 3''', 4''', 5''', 6''', ... 0'''', 1'''', 2'''', 3'''', 4'''', 5'''', 6'''', ... 0''''', 1''''', 2''''', 3''''', 4''''', 5''''', 6''''', ... this gives, starting from up left: 0, 1, 0', 0'', 1', 2, 3, 2', 1'', 0''', 0'''', 1''', 2'', 3', 4, 5, 4', 3'', 2''', 1'''', 0''''', 0'''''', ... and gives an enumeration (bijection with N) of the infinite union NUN'UN''UN'''UN''''U ... Remark: for those who recall the transfinite ordinal we have met when we have try to give a good answer to the first fairy, you can realize that if we put on each set N', N'', N''', etc. the usual order (so that 3''' < 5''') for example, and if we extend those orders on the all infinite union by deciding that x^{n strokes} < y^{m strokes} in case m is bigger than n (whatever are x and y), we get transfinite ordinal number.
Example omega is just N omega+omega is the ordinal of the order: 0, 1, 2, 3, 4, 5, 6, ... 0', 1', 2', 3', 4', 5', 6', ... omega* omega = omega+omega+omega+omega+omega+omega+omega+ ... = the ordinal of the order 0,1,2,3,...0',1',2',3',...0'',1'',2'',3'',...0''',1''',2''',3''',...0''' ',1'''',2'''',3'''',...0''''',1''''',2''''',3''''',...0'''''',1'''''',2' ''''',3'''''',...0''''''',1''''''',2''''''',3''''''', .... Obviously (ask if this is not clear), the zigzagging showing that NUN'UN''UN'''UN''''U ... is in bijection with N, shows that the ordinal omega* omega is in bijection with N. Note that all the "..." in 0,1,2,3,...0',1',2',3',...0'',1'',2'',3'',...0''',1''',2''',3''',...0''' ',1'''',2'''',3'''',...0''''',1''''',2''''',3''''',...0'''''',1'''''',2' ''''',3'''''',...0''''''',1''''''',2''''''',3''''''', .... have the usual interpretation, except the last one, which supposes you have grasped the process for generating that ordinal. (Compare may be with Thorgny's ultrafinitist rebuilding of the ordinals, but only if you grasp well the non-ultrafinitist one before). All transfinite ordinal we have used with the first fairy (to give her a big finite number) were enumerable (= in bijection with N). This ends the little introduction to the notion of bijection, and of (infinite) enumeration (= bijection with N). Next, I will show the existence of bigger infinities, that is, of infinite set which cannot be put in a bijective correspondence with N. You can prepare yourself with the set NXNXNXNXNXNXNXNX ... (the infinite cartesian product of N with itself) This is the set of omega-uples of natural number. It is the set of (x, y, z, r, s, t, u, ...) with x, z, ... all belonging to N. So it is also the set of sequences of natural numbers, and so it is also the set of functions from N to N. or you can take even just, putting 2 for the set {0,1} the infinite product of 2 with itself 2X2X2X2X2X2X2X2X2X2X2X ... (the infinite cartesian product of {0, 1} with itself), This is the set of infinite binary sequences, or the set of functions from N to 2, where 2 is put for {0,1}. I guess many of you knows that such sets are NOT enumerable, and that this can be proved by one application of Cantor's diagonal. But I will explain this in detail. The reason is not that we will have some use of that result, but just because it is the simplest use of a diagonal. Only after that, I will be able to explain, by a similar but a bit subtle diagonal, why Church thesis, and even more generally the hypothesis asserting there exist a universal computing machine, is a quite strong postulate whose impact on the everything theories, the measure problem, etc. is incalculable ... This is needed to have a thorough understanding of the seventh and eighth steps of the UDA. But my main goal long term is to answer precisely a question by David about the origin and importance of the first person in the comp frame. So the sequel is: 1) Cantor's diagonal 2) are there universal computing machine? (Kleene's diagonal, and Church thesis) 3) A fundamental theorem about universal computing machines. (All such machine are imperfect, or insecure) Please ask questions. To miss math due to notation problem is like to miss travels due to mishandling of the use of maps, or to miss love by mishandling of the use of clothes ... It is missing a lot, for mishandling a few I wanna say. Bruno
http://iridia.ulb.ac.be/~marchal/
|
| Re: Bijections (was OM = SIGMA1) |
Torgny Tholerus |
19/11/07 08:00 |
Torgny Tholerus skrev:
If you define the set of all natural numbers N, then you can pull out
the biggest number m from that set. But this number m has a different
"type" than the ordinary numbers. (You see that I have some sort of
"type theory" for the numbers.) The ordinary deduction rules do not
hold for numbers of this new type. For all ordinary numbers you can
draw the conclusion that the successor of the number is included in N.
But for numbers of this new type, you can not draw this conclusion.
You can say that all ordinary natural numbers are of type 0. And the
biggest natural number m, and all numbers you construct from that
number, such that m+1, 2*m, m/2, and so on, are of type 1. And you can
construct a set N1 consisting of all numbers of type 1. In this set
there exists a biggest number. You can call it m1. But this new
number is a number of type 2.
It may look like a contradiction to say that m is included in N, and to
say that all numbers in N have a successor in N, and to say that m have
no successor in N. But it is not a constrdiction because the rule "all
numbers in N have a successor in N" can be expanded to "all numbers of
type 0 in N have a successor in N". And because m is a number of type
1, then that rule is not applicable to m.
You can comapre this with the Russell's paradox. This paradox says:
Construct the set R of all sets that does not contain itself. For this
set R there will be the rule: For all x, if x does not contain itself,
then R contains x.
If we here substitute R for x, then we get: If R does not contain
itself, then R contains R. This is a contradiction.
The contradiction is caused by an illegal conclusion, it is illegal to
substitute R for x in the "For all x"-quantifier above.
This paradox is solved by "type theory". If you say that all ordinary
sets are of type 0, then the set R will be of type 1. And every
all-quantifiers are restricted to objects of a special type. So the
rule above should read: For all x of type 0, if x does not contain
itself, then R contains x.
In this case you will not get any contradiction, because you can not
substitute R for x in that rule.
==========
Compare this with the case of the biggest natural number:
Construct the set N of all natural numbers. For this set N there will
be the rule: For all x, if N contains x, then N contains x+1.
Suppose that there exists a biggest natural number m in N. If we
substitute m for x, then we get: If N contains m, then N contains m+1.
This is a contradiction, because m+1 is bigger than m, so m can not be
the biggest number then.
But the contradiction is caused by an illegal conclusion, it is illegal
to substitute m for x in the "For all x"-quantifier above.
This paradox is solved by "type theory". If you say that all ordinary
natural numbers are of type 0, then the natural number m will be of
type 1. And every all-quantifiers are restricted to objects of a
special type. So the rule above should read: For all x of type 0, if N
contains x, then N contains x+1.
In this case you will not get any contradiction, because you can not
substitute m for x in that rule.
===========
Do you see the similarities in both these cases?
--
Torgny
|
| Re: The big-black-cloud-interpretation. |
Torgny Tholerus |
19/11/07 08:20 |
Torgny Tholerus skrev:
Now you define a new concept INNFINITE, that is defined by:
If you have a bijection from all visible numbers of a set S, to all
visible numbers of a true subset of S, then you say that the set S in
INNFINITE.
Then you can use this concept INNFINITE, and you will get a consistent
theory with no contradictions, because you have a finite visualization
of this theory.
This concept INNFINITY behaves in exact the same way as the concept
infinity in ordinary mathematics. So you do not need the concept
infinity. Every conclusion you do with the concept infinity, you can
do with the concept INNFINITY. So you will not lose anything, if you
discard the concept infinity. Infinity is not needed in mathematics.
--
Torgny
|
| Re: Bijections (was OM = SIGMA1) |
Mirek Dobsicek |
19/11/07 11:14 |
Hi Bruno, thank you for posting the solutions. Of course, I solved it by myself and it was a fine relaxing time to do the paper work trying to be rigorous, however, your solutions gave me additional insights, nice. I am on the board for the sequel. Best, Mirek > > I give the solution of the little exercises on the notion of bijection. >
>
> So the sequel is: > > 1) Cantor's diagonal > 2) are there universal computing machine? (Kleene's diagonal, and > Church thesis) > > 3) A fundamental theorem about universal computing machines. (All such > machine are imperfect, or insecure) > > Please ask questions. To miss math due to notation problem is like to > miss travels due to mishandling of the use of maps, or to miss love by > mishandling of the use of clothes ... It is missing a lot, for > mishandling a few I wanna say. > |
| Re: Bijections (was OM = SIGMA1) |
Bruno Marchal |
20/11/07 01:46 |
Le 19-nov.-07, à 17:00, Torgny Tholerus a écrit : This points on one among many ways to handle Russell's paradox. Type Theories (TT) are nice, but many logicians prefer some untyped set theory, like ZF, or a two types theory like von Neuman Bernays Godel (VBG). Or Cartesian closed categories, toposes, etc. But set theory is a bit out of the scope of this thread. All such theories (ZF, VBG, TT) are example of "Lobian Machine", and my goal is to study all such machine without choosing one in particular, and using traditional math, instead of working really "in" some particular theories. Another solution for many paradoxes consists in working with constructive objects. Soon, this is what we will do, by focusing on the set of "computable functions" instead of the set of all functions. The reason is not to escape paradoxes though. The reason is to learn something about machines (which are finite or constructive object). Just wait a bit. I will first explain Cantor's diagonal, which is simple but rather "transcendental". > Compare this with the case of the biggest natural number: > > Construct the set N of all natural numbers. For this set N there > will be the rule: For all x, if N contains x, then N contains x+1. > > Suppose that there exists a biggest natural number m in N. If we > substitute m for x, then we get: If N contains m, then N contains > m+1. This is a contradiction, because m+1 is bigger than m, so m can > not be the biggest number then. > > But the contradiction is caused by an illegal conclusion, it is > illegal to substitute m for x in the "For all x"-quantifier above. > > This paradox is solved by "type theory". If you say that all > ordinary natural numbers are of type 0, then the natural number m will > be of type 1. And every all-quantifiers are restricted to objects of > a special type. So the rule above should read: For all x of type 0, > if N contains x, then N contains x+1. > > In this case you will not get any contradiction, because you can not > substitute m for x in that rule. > > =========== > > Do you see the similarities in both these cases? > Except that naive usual number theory does not lead to any paradox, unlike naive set theory. *you* got a paradox because of *your* ultrafinistic constraint. So you are proposing a medication which could be worst that the disease I'm afraid. Very few people have any trouble with the potential infinite N = omega = {0, 1, 2, 3, 4, 5, ...}. With comp it can be shown that you don't need more at the ontological third person level. What will happen is that infinities come back in the first person point of views, and are very useful and lawful. Now, sound Lobian Machines can disagree on the real status of some of those infinities, but this does not concern us, at least, not urgently, I think.
Bruno http://iridia.ulb.ac.be/~marchal/
|
| Re: Bijections (was OM = SIGMA1) |
Bruno Marchal |
20/11/07 02:59 |
Hi Mirek, Le 19-nov.-07, à 20:14, Mirek Dobsicek a écrit :
> > Hi Bruno, > > thank you for posting the solutions. Of course, I solved it by myself > and it was a fine relaxing time to do the paper work trying to be > rigorous, however, your solutions gave me additional insights, nice. > > I am on the board for the sequel. Thanks.
I will explain soon (this "afternoon") how Cantor managed to show that some infinite set "have more elements" than the infinite set N, in the sense that there will be no bijection from N to such set, despite obvious bijection between N and subset of such set. I am sure most of you know that proof by diagonal. However, the goal will be to show later how a similar reasoning can put a serious doubt on the existence of universal machine, and on serious constraints such machine have to live with in case we continue to believe in their existence. Before doing that, I want to explain briefly the difference between ordinal and cardinal. This explanation is not necessary for the sequel, but it could help. I will also use the set representation of numbers and ordinals. So I will represent the number 0 by the empty set, and the number n by the set of numbers strictly little than n. So 0 = { }, 1 = {0} = {{}}, 2 = {0, 1} = {{} {{}}}, 3 = {0, 1, 2}, 4 = {0, 1, 2, 3}, ... then omega = N = {0, 1, 2, 3 ...} is the least infinite ordinal. The advantage of such a representation is that "belongness" modelizes the strictly-lesser-than relation, and subsetness modelizes the lesser-than-or equal. I recall that A is a subset of B, if for all x, x belongs to A entails that x belongs to B. In particular for all set A x belongs to A entails x belongs to A, so all sets are subset of themselves. An ordinal is defined by being a linear well founded order. Well-foundness means that all subsets have a least element. The finite ordinal are thus the natural numbers. They all have different cardinals. That is, two different natural numbers (= different finite ordinal) have different cardinality (different "number" of elements). Take 7 and 5, there is no bijection between them, for example. So in the finite realm, ordinal and cardinal coincide. But infinite ordinals can be different, and still have the same cardinality. I have given examples: You can put an infinity of linear well founded order on the set N = {0, 1, 2, 3, ...}. The usual order give the ordinal omega = {0, 1, 2, 3, ...}. Now omega+1 is the set of all ordinal strictly lesser than omega+1, with the convention above. This gives {0, 1, 2, 3, ... omega} = {0, 1, 2, 3, 4, ....{0, 1, 2, 3, 4, ....}}. As an order, and thus as an ordinal, it is different than omega or N. But as a cardinal omega and omega+1 are identical, that means (by definition of cardinal) there is a bijection between omega and omega+1. Indeed, between {0, 1, 2, 3, ... omega} and {0, 1, 2, 3, ...}, you can build the bijection: 0--------omega 1--------0 2--------1 3--------2 ... n ------- n-1 ... All right? "-----" represents a rope. To sum up; finite ordinal and finite cardinal coincide. Concerning infinite "number" there are much ordinals than cardinals. In between two different infinite cardinal, there will be an infinity of ordinal. We have already seen that omega, omega+1, ... omega+omega, omega+omega+1, ....3.omega, ... 4.omega .... ....omega.omega ..... omega.omega.omega, .....omega^omega ..... are all different ordinals, but all have the same cardinality. Don't worry, we will not use that. Question: are there really two different infinite cardinals? That is, are there two infinite sets with different cardinality? That is, are there two different infinite sets A and B without any bijection in between ? The answer is yes, and that is what cantor has discovered by its diagonal construction, and that is the object of the next post. All what I did want to say here, is that automatically, in between A and B, there will be an infinite amount of different ordinals. Bruno
http://iridia.ulb.ac.be/~marchal/ |
| Re: Bijections (was OM = SIGMA1) |
Torgny Tholerus |
20/11/07 03:14 |
Bruno Marchal skrev: > > To sum up; finite ordinal and finite cardinal coincide. Concerning > infinite "number" there are much ordinals than cardinals. In between > two different infinite cardinal, there will be an infinity of ordinal. > We have already seen that omega, omega+1, ... omega+omega, > omega+omega+1, ....3.omega, ... 4.omega .... ....omega.omega ..... > omega.omega.omega, .....omega^omega ..... are all different ordinals, > but all have the same cardinality. >
Was it not an error there? 2^omega is just the number of all subsets of omega, and the number of all subsets always have bigger cardinality than the set. So omega^omega can not have the same cardinality as omega. -- Torgny
|
| Re: Bijections (was OM = SIGMA1) |
Bruno Marchal |
20/11/07 06:41 |
Le 20-nov.-07, à 12:14, Torgny Tholerus a écrit : Yes, that is true.
> So omega^omega can not have the same cardinality as omega. But addition, multiplication, and thus exponentiation are not the same operation for ordinals and cardinals. I should have written omega"^"omega, or something like that. That is why I have written 3.omega instead of 3*omega.
We can come back on ordinal later, but now I will focus the attention on the cardinals, and prove indeed that 2^omega, or 2^N, or equivalently the infinite cartesian product (of sets) 2X2X2X2X2X2X2X2X... , is NOT enumerable (and indeed vastly bigger that the ordinal omega"^"omega. You can look at the thread on the growing functions for a little more on the ordinals. Actually my point was to remind people of the difference between ordinal and cardinal, and, yes, they have different addition, multiplication, etc. Bruno http://iridia.ulb.ac.be/~marchal/ |
| Cantor's Diagonal |
Bruno Marchal |
20/11/07 08:40 |
Hi, David, are you still there? This is a key post, with respect to the "Church Thesis" thread. So let us see that indeed there is no bijection between N and 2^N = 2X2X2X2X2X2X... = {0,1}X{0,1}X{0,1}X{0,1}X... = the set of infinite binary sequences. Suppose that there is a bijection between N and the set of infinite binary sequences. Well, it will look like that, where again "----" represents the "ropes": 0 -------------- (1, 0, 0, 1, 1, 1, 0 ... 1 -------------- (0, 0, 0, 1, 1, 0, 1 ... 2 --------------- (0, 1, 0, 1, 0, 1, 1, ... 3 --------------- (1, 1, 1, 1, 1, 1, 1, ... 4 --------------- (0, 0, 1, 0, 0, 1, 1, ... 5 ----------------(0, 0, 0, 1, 1, 0, 1, ... ... My "sheep" are the natural numbers, and my neighbor's sheep are the infinite binary sequences (the function from N to 2, the elements of the infinite cartesian product 2X2X2X2X2X2X... ). My flock of sheep is the *set* of natural numbers, and my neighbor's flock of sheep is the *set* of all infinite binary sequences. Now, if this: 0 -------------- (1, 0, 0, 1, 1, 1, 0 ... 1 -------------- (0, 0, 0, 1, 1, 0, 1 ... 2 --------------- (0, 1, 0, 1, 0, 1, 1, ... 3 --------------- (1, 1, 1, 1, 1, 1, 1, ... 4 --------------- (0, 0, 1, 0, 0, 1, 1, ... 5 ----------------(0, 0, 0, 1, 1, 0, 1, ... ... is really a bijection, it means that all the numbers 1 and 0 appearing on the right are well determined (perhaps in Platonia, or in God's mind, ...). But then the diagonal sequence, going from the left up to right down, and build from the list of binary sequences above: 1 0 0 1 0 0 ... is also completely well determined (in Platonia or in the mind of a God). But then the complementary sequence (with the 0 and 1 permuted) is also well defined, in Platonia or in the mind of God(s) 0 1 1 0 1 1 ... But this infinite sequence cannot be in the list, above. The "God" in question has to ackonwledge that. The complementary sequence is clearly different -from the 0th sequence (1, 0, 0, 1, 1, 1, 0 ..., because it differs at the first (better the 0th) entry. -from the 1th sequence (0, 0, 0, 1, 1, 0, 1 ... because it differs at the second (better the 1th) entry. -from the 2th sequence (0, 0, 0, 1, 1, 0, 1 ... because it differs at the third (better the 2th) entry. and so one. So, we see that as far as we consider the bijection above well determined (by God, for example), then we can say to that God that the bijection misses one of the neighbor sheep, indeed the "sheep" constituted by the infinite binary sequence complementary to the diagonal sequence cannot be in the list, and that sequence is also well determined (given that the whole table is). But this means that this bijection fails. Now the reasoning did not depend at all on the choice of any particular bijection-candidate. Any conceivable bijection will lead to a well determined infinite table of binary numbers. And this will determine the diagonal sequence and then the complementary diagonal sequence, and this one cannot be in the list, because it contradicts all sequences in the list when they cross the diagonal (do the drawing on paper). Conclusion: 2^N, the set of infinite binary sequences, is not enumerable. All right? Next I will do again that proof, but with notations instead of drawing, and I will show more explicitly how the contradiction arise. Exercice-training: show similarly that N^N, the set of functions from N to N, is not enumerable.
Bruno http://iridia.ulb.ac.be/~marchal/
|
| Re: Cantor's Diagonal |
David Nyman |
20/11/07 08:47 |
On 20/11/2007, Bruno Marchal <mar...@ulb.ac.be> wrote: > David, are you still there? This is a key post, with respect to the > "Church Thesis" thread. Sorry Bruno, do forgive me - we seem destined to be out of synch at the moment. I'm afraid I'm too distracted this week to respond adequately - back on-line next week at the latest. David |
| Re: Bijections (was OM = SIGMA1) |
Brent |
20/11/07 08:59 |
Bruno Marchal wrote: > . > > But infinite ordinals can be different, and still have the same > cardinality. I have given examples: You can put an infinity of linear > well founded order on the set N = {0, 1, 2, 3, ...}. What is the definition of "linear well founded order"? I'm familiar with "well ordered", but how is "linear" applied to sets? Just curious. Brent Meeker
|
| Re: Cantor's Diagonal |
Torgny Tholerus |
20/11/07 09:42 |
Bruno Marchal skrev:
But then the complementary sequence (with the 0 and 1
permuted) is
also well defined, in Platonia or in the mind of God(s)
0 1 1 0
1 1 ...
But this infinite sequence cannot be in the list, above.
The "God" in question has to ackonwledge that.
The complementary sequence is clearly different
-from the 0th sequence (1, 0, 0,
1, 1, 1, 0 ..., because it differs at the first (better the 0th) entry.
-from the 1th sequence (0, 0, 0,
1, 1, 0, 1 ... because it differs at the second (better the 1th) entry.
-from the 2th sequence (0, 0, 0,
1, 1, 0, 1 ... because it differs at the third (better the 2th) entry.
and so one.
So, we see that as far as we consider the bijection above well
determined (by God, for example), then we can say to that God that the
bijection misses one of the neighbor sheep, indeed the "sheep"
constituted by the infinite binary sequence complementary to the
diagonal sequence cannot be in the list, and that sequence is also
well determined (given that the whole table is).
But this means that this bijection fails. Now the reasoning did not
depend at all on the choice of any particular bijection-candidate. Any
conceivable bijection will lead to a well determined infinite
table of binary numbers. And this will determine the diagonal sequence
and then the complementary diagonal sequence, and this one cannot be
in the list, because it contradicts all sequences in the list when
they cross the diagonal (do the drawing on paper).
Conclusion: 2^N, the set of infinite binary sequences, is not
enumerable.
All right?
An ultrafinitist comment to this:
======
You can add this complementary sequence to the end of the list. That
will make you have a list with this complementary sequence included.
But then you can make a new complementary sequence, that is not
inluded. But you can then add this new sequence to the end of the
extended list, and then you have a bijection with this new sequence
also. And if you try to make another new sequence, I will add that
sequence too, and this I will do an infinite number of times. So you
will not be able to prove that there is no bijection...
======
What is wrong with this conclusion?
--
Torgny
|
| Re: Bijections (was OM = SIGMA1) |
Torgny Tholerus |
20/11/07 10:01 |
Bruno Marchal skrev: > > But infinite ordinals can be different, and still have the same > cardinality. I have given examples: You can put an infinity of linear > well founded order on the set N = {0, 1, 2, 3, ...}. > The usual order give the ordinal omega = {0, 1, 2, 3, ...}. Now omega+1 > is the set of all ordinal strictly lesser than omega+1, with the > convention above. This gives {0, 1, 2, 3, ... omega} = {0, 1, 2, 3, 4, > ....{0, 1, 2, 3, 4, ....}}. As an order, and thus as an ordinal, it is > different than omega or N. But as a cardinal omega and omega+1 are > identical, that means (by definition of cardinal) there is a bijection > between omega and omega+1. Indeed, between {0, 1, 2, 3, ... omega} and > {0, 1, 2, 3, ...}, you can build the bijection: > > 0--------omega > 1--------0 > 2--------1 > 3--------2 > ... > n ------- n-1 > ... > > All right? "-----" represents a rope. >
An ultrafinitist comment: In the last line of this sequence you will have: ? --------- omega-1 But what will the "?" be? It can not be omega, because omega is not included in N... -- Torgny
|
| Re: Cantor's Diagonal |
Brent |
20/11/07 10:27 |
Torgny Tholerus wrote: > Bruno Marchal skrev: >> But then the complementary sequence (with the 0 and 1 permuted) is >> also well defined, in Platonia or in the mind of God(s) >>
>> *0* *1* *1* *0* *1* *1* ... >> >> But *this* infinite sequence cannot be in the list, above. The "God" >> in question has to ackonwledge that. >> The complementary sequence is clearly different
>> -from the 0th sequence (*_1_*, 0, 0, 1, 1, 1, 0 ..., because it >> differs at the first (better the 0th) entry.
>> -from the 1th sequence (0, *_0_*, 0, 1, 1, 0, 1 ... because it >> differs at the second (better the 1th) entry.
>> -from the 2th sequence (0, 0, *_0_*, 1, 1, 0, 1 ... because it You'd have to insert the new sequence in the beginning, as there is no "end of the list". Brent Meeker > > -- > Torgny > > > |
| Re: Cantor's Diagonal |
Barry Brent |
20/11/07 14:39 |
You're saying that, just because you can *write down* the missing sequence (at the beginning, middle or anywhere else in the list), it follows that there *is* no missing sequence. Looks pretty wrong to me. Cantor's proof disqualifies any candidate enumeration. You respond by saying, "well, here's another candidate!" But Cantor's procedure disqualified *any*, repeat *any* candidate enumeration. Barry Brent Dr. Barry Brent barry...@earthlink.net http://home.earthlink.net/~barryb0/ |
| Re: Cantor's Diagonal |
Torgny Tholerus |
20/11/07 23:49 |
meekerdb skrev:
Torgny Tholerus wrote:
An ultrafinitist comment to this:
======
You can add this complementary sequence to the end of the list. That
will make you have a list with this complementary sequence included.
But then you can make a new complementary sequence, that is not
inluded. But you can then add this new sequence to the end of the
extended list, and then you have a bijection with this new sequence
also. And if you try to make another new sequence, I will add that
sequence too, and this I will do an infinite number of times. So you
will not be able to prove that there is no bijection...
======
What is wrong with this conclusion?
You'd have to insert the new sequence in the beginning, as there is no
"end of the list".
Why can't you add something to the end of the list? In an earlier
message Bruno wrote:
"Now omega+1 is the set of all ordinal strictly lesser than omega+1,
with the convention above. This gives {0, 1, 2, 3, ... omega} = {0, 1,
2, 3, 4, ....{0, 1, 2, 3, 4, ....}}."
In this sentence he added omega to the end of the list of natural
numbers...
--
Torgny
|
| Re: Cantor's Diagonal |
Bruno Marchal |
21/11/07 07:45 |
Le 20-nov.-07, à 17:47, David Nyman a écrit : > > On 20/11/2007, Bruno Marchal <mar...@ulb.ac.be> wrote: > >> David, are you still there? This is a key post, with respect to the >> "Church Thesis" thread. > > Sorry Bruno, do forgive me - we seem destined to be out of synch at > the moment. I'm afraid I'm too distracted this week to respond > adequately - back on-line next week at the latest. Take it easy. It is the main advantage of electronical conversation, unlike the phone, we can answer at the best moment. But I appreciate you tell me.
"See" you next week. Bruno PS I'm afraid I have not the time to comment the last posts today, except for elementary question perhaps, but I will have more time soon (probably or hopefully tomorrow). http://iridia.ulb.ac.be/~marchal/ |
| Re: Bijections (was OM = SIGMA1) |
Bruno Marchal |
21/11/07 07:48 |
Le 20-nov.-07, à 17:59, meekerdb a écrit : By linear, I was just meaning a non branching order. A tree can be well founded too, meaning all its branches have a "length" given by an ordinal.
Bruno http://iridia.ulb.ac.be/~marchal/
|
| Re: Cantor's Diagonal |
Bruno Marchal |
21/11/07 08:00 |
Le 20-nov.-07, à 23:39, Barry Brent wrote : > > You're saying that, just because you can *write down* the missing > sequence (at the beginning, middle or anywhere else in the list), it > follows that there *is* no missing sequence. Looks pretty wrong to me. > > Cantor's proof disqualifies any candidate enumeration. You respond > by saying, "well, here's another candidate!" But Cantor's procedure > disqualified *any*, repeat *any* candidate enumeration. > > Barry Brent Torgny, I do agree with Barry. Any bijection leads to a contradiction, even in some effective way, and that is enough (for a classical logician).
But look what you write:
> On Nov 20, 2007, at 11:42 AM, Torgny Tholerus wrote: > >> >> An ultrafinitist comment to this: >> ====== >> You can add this complementary sequence to the end of the list. >> That will make you have a list with this complementary sequence >> included. >> >> But then you can make a new complementary sequence, that is not >> inluded. But you can then add this new sequence to the end of the >> extended list, and then you have a bijection with this new sequence >> also. And if you try to make another new sequence, I will add that >> sequence too, and this I will do an infinite number of times. How could an ultrafinitist refute an argument by saying "... and this I will do an infinite number of times. "?
>> So >> you will not be able to prove that there is no bijection...
Actually no. If you do what you described omega times, you will just end up with a set which can still be put in 1-1 correspondence with N (as shown in preceding posts on bijections) To refute Cantor, here, you should do what you described a very big infinity of times, indeed an non enumerable infinity of times. But then you have to assume the existence of a non enumerable set at the start. OK?
Bruno http://iridia.ulb.ac.be/~marchal/ >> ====== >> What is wrong with this conclusion? >> >> -- >> Torgny >> >>> > > Dr. Barry Brent > barry...@earthlink.net > http://home.earthlink.net/~barryb0/ > > > > > > >
|
| Re: Cantor's Diagonal |
Bruno Marchal |
21/11/07 08:18 |
Le 21-nov.-07, à 08:49, Torgny Tholerus a écrit : Adding something to the end or to the middle or to the beginning of an infinite list, does not change the cardinality of that list. And in Cantor proof, we are interested only in the cardinality notion.
Adding something to the beginning or to the end of a infinite ORDERED list, well, it does not change the cardinal of the set involved, but it obviosuly produce different order on those sets, and this can give different ordinal, which denote type of order (isomorphic order). The ordered set {0, 1, 2, 3, ...} has the same cardinality that the ordered set {1, 2, 3, 4, ... 0} (where by definition 0 is bigger than all natural numbers). But they both denote different ordinal, omega, and omega+1 respectively. Note that {1, 0, 2, 3, 4, ...} is a different order than {0, 1, 2, 3, ...}, but both order here are isomorphic, and correspond to the same ordinal (omega). That is why 1+omega = omega, and omega+1 is different from omega. Adding one object in front of a list does not change the type of the order. Adding an element at the end of an infinite list does change the type of the order. {0, 1, 2, 3, ...} has no bigger element, but {1, 2, 3, ... 0} has a bigger element. So, you cannot by simple relabelling of the elements get the same type of order (and thus they correspond to different ordinals). OK? (this stuff will not be used for Church Thesis, unless we go very far ...later). Bruno
http://iridia.ulb.ac.be/~marchal/
|
| Re: Cantor's Diagonal |
Torgny Tholerus |
21/11/07 08:33 |
Bruno Marchal skrev:
Le 20-nov.-07, à 23:39, Barry Brent wrote :
You're saying that, just because you can *write down* the missing
sequence (at the beginning, middle or anywhere else in the list), it
follows that there *is* no missing sequence. Looks pretty wrong to me.
Cantor's proof disqualifies any candidate enumeration. You respond
by saying, "well, here's another candidate!" But Cantor's procedure
disqualified *any*, repeat *any* candidate enumeration.
Barry Brent
Torgny, I do agree with Barry. Any bijection leads to a contradiction,
even in some effective way, and that is enough (for a classical
logician).
What do you think of this "proof"?:
Let us have the bijection:
0 -------- {0,0,0,0,0,0,0,...}
1 -------- {1,0,0,0,0,0,0,...}
2 -------- {0,1,0,0,0,0,0,...}
3 -------- {1,1,0,0,0,0,0,...}
4 -------- {0,0,1,0,0,0,0,...}
5 -------- {1,0,1,0,0,0,0,...}
6 -------- {0,1,1,0,0,0,0,...}
7 -------- {1,1,1,0,0,0,0,...}
8 -------- {0,0,0,1,0,0,0,...}
...
omega --- {1,1,1,1,1,1,1,...}
What do we get if we apply Cantor's Diagonal to this?
--
Torgny
|
| Theory of Everything based on E8 by Garrett Lisi |
George Levy |
21/11/07 10:54 |
A theory of everyting is sweeping the Physics community.
The theory by Garrett Lisi is explained in this Wiki
entry.
A simulation of E8 can be found a the
New Scientist.
The Wiki
entry on E8 is also interesting.
George
|
| Re: Cantor's Diagonal |
Barry Brent |
21/11/07 14:14 |
|
| elaboration Re: Cantor's Diagonal |
Barry Brent |
21/11/07 22:19 |
The reason it isn't a bijection (of a denumerable set with the set of binary sequences): the pre-image (the left side of your map) isn't a set--you've imposed an ordering. Sets, qua sets, don't have orderings. Orderings are extra. (I'm not a specialist on this stuff but I think Bruno, for example, will back me up.) It must be the case that you won't let us identify the left side, for example, with {omega, 0, 1, 2, ... }, will you? For if you did, it would fall under Cantor's argument. Barry On Nov 21, 2007, at 10:33 AM, Torgny Tholerus wrote: Dr. Barry Brent barry...@earthlink.net http://home.earthlink.net/~barryb0/ |
| Re: Theory of Everything based on E8 by Garrett Lisi |
Bruno Marchal |
22/11/07 04:10 |
Le 21-nov.-07, à 19:54, George Levy a écrit : Thanks, very interesting indeed. Note that the original paper is accessible from the New Scientist entry. Not so easy to read (need of differential geometry, simple groups, etc. Quite close to the idea of the importance of 24 which I mention periodically ... :)
Now such work raises the remark, which I don't really want to develop now, which is that qualifiying "TOE" a theory explaining "only" forces and particles or field, is implicit physicalism, and we know (by UDA) that this is incompatible with comp. Yet I bet Lisi is quite close to the sort of physics derivable by machine's or number's introspection. Actually, getting physics from so "few" symmetries is a bit weird (I have to study the paper in detail). With comp, we have to explain the symmetries *and* the geometry, and the quantum logic, from the numbers and their possible stable discourses ... If not, it is not a theory of everything, but just a classification, a bit like the Mendeleev table classifies atoms without really explaining. But Lisi's theory seems beautiful indeed ... Bruno http://iridia.ulb.ac.be/~marchal/
|
| Re: elaboration Re: Cantor's Diagonal |
Bruno Marchal |
22/11/07 04:17 |
Le 22-nov.-07, à 07:19, Barry Brent a écrit : > > The reason it isn't a bijection (of a denumerable set with the set of > binary sequences): the pre-image (the left side of your map) isn't > a set--you've imposed an ordering. Sets, qua sets, don't have > orderings. Orderings are extra. (I'm not a specialist on this stuff > but I think Bruno, for example, will back me up.) It must be the > case that you won't let us identify the left side, for example, with > {omega, 0, 1, 2, ... }, will you? For if you did, it would fall under > Cantor's argument. I agree. Presently, I prefer not talking too much on the ordinals, because it could be confusing for many. More later ...
Bruno http://iridia.ulb.ac.be/~marchal/ |
| Re: Cantor's Diagonal |
Bruno Marchal |
22/11/07 06:14 |
Le 21-nov.-07, à 17:33, Torgny Tholerus a écrit : > What do you think of this "proof"?: > > Let us have the bijection: > > 0 -------- {0,0,0,0,0,0,0,...} > 1 -------- {1,0,0,0,0,0,0,...} > 2 -------- {0,1,0,0,0,0,0,...} > 3 -------- {1,1,0,0,0,0,0,...} > 4 -------- {0,0,1,0,0,0,0,...} > 5 -------- {1,0,1,0,0,0,0,...} > 6 -------- {0,1,1,0,0,0,0,...} > 7 -------- {1,1,1,0,0,0,0,...} > 8 -------- {0,0,0,1,0,0,0,...} > ... > omega --- {1,1,1,1,1,1,1,...} > > What do we get if we apply Cantor's Diagonal to this? Note also that in general, we start from what we want to prove, and then do the math. Your idea of transfinite (ordinal) diagonalisation is cute though, but I have currently no idea where this could lead. BTW, it is also funny that such a transfinite idea is proposed by an ultrafinistist!
I guess you have seen that {(0,0,0,0,0,0,0,...), (1,0,0,0,0,0,0,...), ... does clearly not enumerate the infinite sequences (you don't have to use the diagonal for showing that. It is also better to use parentheses instead of accolades, given that the binary sequences are ordered (notation detail). Bruno http://iridia.ulb.ac.be/~marchal/
|
| Re: Are First Person prime? |
George Levy |
22/11/07 11:50 |
Hi Bruno,
I am reopening an old thread ( more than a year old) which I found very
intriguing. It leads to some startling conclusions.
Le 05-août-06, à 02:07, George Levy a écrit :
Bruno Marchal wrote: I think that if you want to
make the first person primitive, given that neither you
nor
me can
really define it, you will need at least to axiomatize it in some way.
Here is my question. Do you agree that a first person is a knower, and
in that case, are you willing to accept the traditional axioms for
knowing. That is:
1) If p is knowable then p is true;
2) If p is knowable then it is knowable that p is knowable;
3) if it is knowable that p entails q, then if p is knowable then q is
knowable
(+ some logical rules).
Bruno, what or who do you mean by "it" in statements 2) and 3). In
addition, what do you mean by "is knowable", "is true" and "entails"?
Are "is knowable", "is true" and "entails" absolute or do they have
meaning only with respect to a particular observer? Can these terms be
relative to an observer? If they can, how would you rephrase these
statements?
George
|
| Re: Are First Person prime? |
George Levy |
22/11/07 11:56 |
One more question: can or should p be the observer?
George
|
| Re: Theory of Everything based on E8 by Garrett Lisi |
marc.geddes |
22/11/07 18:30 |
On Nov 23, 1:10 am, Bruno Marchal <marc...@ulb.ac.be> wrote:
>
> Now such work raises the remark, which I don't really want to develop
> now, which is that qualifiying "TOE" a theory explaining "only" forces
> and particles or field, is implicit physicalism, and we know (by UDA)
> that this is incompatible with comp.
Yes indeed Bruno. As far as I tell tell, all of physics is ultimately
geometry. But as we've pointed out on this list many times, a theory
of physics is *not* a theory of everything, since it makes the
(probably false) assumption that everything is reducible to physical
substances and properties. Thus we both are in agreement on this, but
for different reasons (you because, you think math is the ultimate
basis of everything aka COMP, me, because of my property dualism, aka
the need for a triple-aspect explanation of physical/teleological/
mathematical properties as the basis for everything).
We keep telling mainstream scients, but mainstream scients are not
listening to us. *sigh*.
> Yet I bet Lisi is quite close to the sort of physics derivable by
> machine's or number's introspection. Actually, getting physics from so
> "few" symmetries is a bit weird (I have to study the paper in detail).
> With comp, we have to explain the symmetries *and* the geometry, and
> the quantum logic, from the numbers and their possible stable
> discourses ... If not, it is not a theory of everything, but just a
> classification, a bit like the Mendeleev table classifies atoms without
> really explaining. But Lisi's theory seems beautiful indeed ...
>
> Bruno
>
There's too many people mucking around with physics - I do wish more
people were working on computer science. Physics is the most advanced
of our sciences, but computer science lags behind. It just seems to
be an unfortunate historical accident that physical theories developed
first and then lots of social status got attached to theoretical
physics (stemming from the glorification of Newton in Europe).
As a result, physics has advanced well ahead of comp-sci, and there's
lots of money and status attached to physics breakthroughs. But comp-
sci is actually far more important to us in practical sense -
artificial general intelligence would be way way more valuable than
any fundamental physics breakthrough. We would have had real AGI long
ago if there was the same money and glory for comp-sci as there is for
physics *sigh*.
|
| Re: Theory of Everything based on E8 by Garrett Lisi |
Torgny Tholerus |
22/11/07 23:49 |
marc....@gmail.com skrev: > > As far as I tell tell, all of physics is ultimately > geometry. But as we've pointed out on this list many times, a theory > of physics is *not* a theory of everything, since it makes the > (probably false) assumption that everything is reducible to physical > substances and properties.
I think that everything is reducible to physical substances and properties. And I think that all of physics is reducible to pure mathematics... I have now read Garrett Lisis paper. It was interesting, but it is still to early to say if it is important. There is a lot of symmetries in the elementary particles, and there is a lot of symmetries in the E8 Lie group. So it is not any suprise that they both can be mapped on each other. Lisi has mapped 222 elementary particles on the 242 elements of E8, and he has predicted that the rest of the 20 elements correspond to 20 yet to be discovered elementary particles. If it is true, then Lisi will have the Nobel price. If it is not, then we will have to look for another TOE. But it is possible that we will never find any TOE. Because there is 10^500 different possiblities for our universe, and all of these 10^500 universes exist in the same way. By experiments we will have to decide which of those that is our universe, but we will never reach the correct answer, the number of experiments needed will be too many. -- Torgny
|
| RE: Theory of Everything based on E8 by Garrett Lisi |
rafa |
23/11/07 00:19 |
Sorry, but I think Lisi's paper is fatally flawed. Adding altogether fermions and bosons is plain wrong. Best
> Date: Thu, 22 Nov 2007 18:30:03 -0800 > Subject: Re: Theory of Everything based on E8 by Garrett Lisi > From: marc....@gmail.com> To: everyth...@googlegroups.com |
| Re: Are First Person prime? |
Bruno Marchal |
23/11/07 06:53 |
Le 22-nov.-07, à 20:50, George Levy a écrit : > Hi Bruno, > > I am reopening an old thread ( more than a year old) which I found > very intriguing. It leads to some startling conclusions. > > Le 05-août-06, à 02:07, George Levy a écrit : > > Bruno Marchal wrote:I think that if you want to >> make the first person primitive, given that neither you nor me can >> really define it, you will need at least to axiomatize it in some >> way. >> Here is my question. Do you agree that a first person is a knower, >> and >> in that case, are you willing to accept the traditional axioms for >> knowing. That is: >> >> 1) If p is knowable then p is true; >> 2) If p is knowable then it is knowable that p is knowable; >> 3) if it is knowable that p entails q, then if p is knowable then q >> is >> knowable >> >> (+ some logical rules). >> > > Bruno, what or who do you mean by "it" in statements 2) and 3). The same as in "it is raining". I could have written 1. and 2. like
1) knowable(p) -> p 2) knowable(p) -> knowable(knowable(p)) In this way we can avoid using words like "it", or even like "true". "p" is a variable, and is implicitly universally quantified over. "knowable(p) -> p" really means that whatever is the proposition p, if it is knowable then it is true. The false is unknowable (although it could be conceivable, believable, even provable (in inconsistent theory), etc. The "p" in 1. 2. and 3. is really like the "x" in the formula (sin(x))^2 + (cos(x))^2 = 1. "knowable(p) -> p" really means that we cannot know something false. This is coherent with the natural language use of know, which I illustrate often by remarking that we never say "Alfred knew the earth is flat, but the he realized he was wrong". We say instead "Alfred believed that earth is flat, but then ...". . The axiom 1. is the incorrigibility axiom: we can know only the truth. Of course we can believe we know something until we know better. The axiom 2. is added when we want to axiomatize a notion of knowledge from the part of sufficiently introspective subject. It means that if some proposition is knowable, then the knowability of that proposition is itself knowable. It means that when the subject knows some proposition then the subject will know that he knows that proposition. The subject can know that he knows. > In addition, what do you mean by "is knowable", "is true" and > "entails"? All the point in axiomatizing some notion, consists in giving a way to reason about that notion without ever defining it. We just try to agree on some principles, like 1.,2., 3., and then derives things from those principles. Nuance can be added by adding new axioms if necessary. Of course axioms like above are not enough, we have to use deduction rules. In case of the S4 theory, which I will rewrite with modal notation (hoping you recognize it). I write Bp for B(p) to avoid heaviness in the notation, likewize, I write BBp for B(B(p)). 1) Bp -> p (incorrigibility) 2) Bp -> BBp (introspective knowledge) 3) B(p->q) -> (Bp -> Bq) (weak omniscience, = knowability of the consequences of knowable propositions). Now with such axioms you can derive no theorems (except the axiom themselves). So you need some principles which give you a way to deduce theorems from axioms. The usual deduction rule of S4 are the substitution rule, the modus ponens rule and the necessitation rule. The substitution rule say that you can substitute p by any proposition (as far as you avoid clash of variable, etc.). The modus ponens rule say that if you have already derived some formula A, and some formula A -> B, then you can derive B. The necessitation rule says that if you have already derive A, then you can derive BA. > Are "is knowable", "is true" and "entails" absolute or do they have > meaning only with respect to a particular observer? The abstract S4 theory is strictly neutral on this. But abstract theory can have more concrete models or interpretations. In our lobian setting, it will happen that "formal provability by a machine" does not obey the incorrigibility axiom (as Godel notices in his 1933 paper). This means that formal provability by a machine cannot be used to modelize the knowability of the machine. It is a bit counterintuitive, but formal provability by a machine modelizes only a form of "opinion" by the machine, so that to get a knowability notion from the provability notion, we have to explicitly define knowability(p) by "provability(p) and p is true". (Cf Platos's Theaetetus). Here provability and knowability is always relative to an (ideal) machine. I will come back on this in my explanation to David later. But don't hesitate to ask question before.
> Can these terms be relative to an observer? If they can, how would you > rephrase these statements?
An observer ? I guess you mean a subject. Observability could obeys quite different axioms that knowability (as it is the case for machine with comp). Just interpret "knowable(p)" by "p is knowable by M". "M" denotes some machine or entity belonging to some class of machine/entity in which we are interested.
> One more question: can or should p be the observer? "p" has to refer to a proposition. Of course in english (at least in french) we often use similar word with different denotation or meaning. For example, you can say "I know Paul". And Paul, a priori is not a poposition. But such a "know" is not the same as in "I know that Paul is a good guy". S4 is a good candidate for the second "know". The "know" (in "I know Paul") has a quite different meaning, somehow out of topic (to be short). Actually "I know Paul" really means humans variate and pragmatic things like "I met Paul before", or "I know Paul is not the right guy to hire for this job", etc.
With the epistemological sense of "knowing", we cannot know a knower, nor an observer. We can only know propositions. Those proposition copuld bear on a knower: like in I know that Paul know that 17 is prime. Sure. Of course we can observe an observer. This illustarte already that observations and knowledge obeys different logics; hopefully related, of course, as it is with the arithmetical hypostases). Oops, I must already go. Have a good week-end, George, and all of you, Bruno
PS Marc, Thorgny: I will comment your post Monday or Tuesday. http://iridia.ulb.ac.be/~marchal/
|
| Re: Are First Person prime? |
George Levy |
24/11/07 12:23 |
Bruno thank you for this elaborate reply. I would like these three
statements to make use of cybernetic language, that is to be more
explicit in terms of the machine or entity to which they refer. Would
it be correct to rephrase the statements in the active tense, using the
machine as the subject, replacing proposition p by the term data and
replacing "true" by "exist"? The statements would then be:
In a world W there is a machine M, data p and data q such that
1) If M has access to p (possibly in its memory), then p exists in W.
2) If M has access to p, then M has access to the access point to p.
3) If M has access to the information relating or linking p to q then
if M has access to p, it also has access to q.
I assumed that the term "has access" means "in its memory"... but it
does not have to.
I also assumed in statements 3 that the multiple uses of M refers to
the same machine. I guess there may be cases where multiple machines
can have access to the dame data.
Same with statement 4
George
|
| Re: Theory of Everything based on E8 by Garrett Lisi |
marc.geddes |
25/11/07 19:17 |
On Nov 23, 8:49 pm, Torgny Tholerus < tor...@dsv.su.se> wrote:
> marc.ged...@gmail.com skrev:
>
>
>
> > As far as I tell tell, all of physics is ultimately
> > geometry. But as we've pointed out on this list many times, a theory
> > of physics is *not* a theory of everything, since it makes the
> > (probably false) assumption that everything is reducible to physical
> > substances and properties.
>
> I think that everything is reducible to physical substances and
> properties. And I think that all of physics is reducible to pure
> mathematics...
You can't have it both ways. If physics was reducible to pure
mathematics, then physics could not be the 'ontological base level' of
reality and hence everything could not be expressed solely in terms of
physical substance and properties.
Besides which, mathematics and physics are dealing with quite
different distinctions. It is a 'type error' it try to reduce or
identity one with the other.
Mathematics deals with logical properties, physics deals with spatial
(geometric) properties. Although geometry is thought of as math, it
is actually a branch of physics, since in addition to pure logical
axioms, all geometry involves 'extra' assumptions or axioms which are
actually *physical* in nature (not purely mathematical) .
|
| Re: Theory of Everything based on E8 by Garrett Lisi |
Torgny Tholerus |
26/11/07 01:09 |
rafael jimenez buendia skrev:
Sorry, but I think Lisi's paper is fatally flawed. Adding
altogether fermions and bosons is plain wrong. Best
What is wrong with adding fermions and bosons together? Xiao-Gang Wen
is working with a condensed string-net where the waves behave just like
bosons (fotons) and the end of the open strings behave just like
fermions (electrons).
--
Torgny Tholerus
|
| Re: Theory of Everything based on E8 by Garrett Lisi |
Torgny Tholerus |
26/11/07 01:51 |
marc....@gmail.com skrev:
On Nov 23, 8:49 pm, Torgny Tholerus <tor...@dsv.su.se> wrote:
I think that everything is reducible to physical substances and
properties. And I think that all of physics is reducible to pure
mathematics...
You can't have it both ways. If physics was reducible to pure
mathematics, then physics could not be the 'ontological base level' of
reality and hence everything could not be expressed solely in terms of
physical substance and properties.
Besides which, mathematics and physics are dealing with quite
different distinctions. It is a 'type error' it try to reduce or
identity one with the other.
Mathematics deals with logical properties, physics deals with spatial
(geometric) properties. Although geometry is thought of as math, it
is actually a branch of physics, since in addition to pure logical
axioms, all geometry involves 'extra' assumptions or axioms which are
actually *physical* in nature (not purely mathematical) .
When I talk about "pure mathematics" I mean that kind of mathematics
you have in GameOfLife. There you have "gliders" that move in the
GameOfLife-universe, and these gliders interact with eachother when
they meet. These gliders you can see as physical objects. These
physical objects are reducible to pure mathematics, they are the
consequences of the rules behind GameOfLife.
--
Torgny
|
| Re: Are First Person prime? |
Bruno Marchal |
26/11/07 06:01 |
George, you can do that indeed, but then you are particularizing things. This can be helpful from a pedagogical point of view, but the advantage of the axiomatic approach (to a knowledge theory) is that once you agree on the axioms and rules, then you agree on the consequences independently of the particular instantiation you think about. Word like "machine", "access", "memory", "world", data, are, fundamentally harder than the simple idea of knowledge the modal S4 axioms convey. Using machines, for example, could seem as a computationalist restriction, when the axioms S4 remains completely neutral, etc. Also, acceding a memory is more "opinion" than knowledge because we can have false memory for example. (And then what are the inference rules of your system?). S4 is a normal modal logic with natural Kripke referentials (transitive, reflexive accessibility relations). A bit more problematic is your identification of "true" with "exist". This hangs on possible but highly debatable and complex relations between truth and reality. This is interesting per se, but imo a bit out of topics, or premature (in current thread). Perhaps we will have opportunity to debate on this, but I want make sure that what I am explaining now does not depend on those possible relations (between truth and reality). Bruno Le 24-nov.-07, à 21:23, George Levy a écrit :
>> "p" has to refer to a proposition. Of course in english (at least in >> french) we often use similar word with different denotation or >> meaning. For example, you can say "I know Paul". And Paul, a priori >> is not a poposition. But such a "know" is not the same as in "I know >> that Paul is a good guy". >> S4 is a good candidate for the second "know". The "know" (in "I know >> Paul") has a quite different meaning, somehow out of topic (to be >> short). Actually "I know Paul" really means humans variate and >> pragmatic things like "I met Paul before", or "I know Paul is not the >> right guy to hire for this job", etc. >> >> With the epistemological sense of "knowing", we cannot know a >> knower, nor an observer. We can only know propositions. Those >> proposition copuld bear on a knower: like in I know that Paul know >> that 17 is prime. Sure. >> Of course we can observe an observer. This illustarte already that >> observations and knowledge obeys different logics; hopefully related, >> of course, as it is with the arithmetical hypostases). >> >> Oops, I must already go. Have a good week-end, George, and all of >> you, >> >> >> Bruno >> >> PS Marc, Thorgny: I will comment your post Monday or Tuesday. >> >> >> >> >> http://iridia.ulb.ac.be/~marchal/ >> >> >> > > > > >
http://iridia.ulb.ac.be/~marchal/ |
| Re: Theory of Everything based on E8 by Garrett Lisi |
Bruno Marchal |
26/11/07 06:54 |
Le 26-nov.-07, à 04:17, marc....@gmail.com a écrit : > > > > On Nov 23, 8:49 pm, Torgny Tholerus <tor...@dsv.su.se> wrote: >> marc.ged...@gmail.com skrev: >> >> >> >>> As far as I tell tell, all of physics is ultimately >>> geometry. But as we've pointed out on this list many times, a theory >>> of physics is *not* a theory of everything, since it makes the >>> (probably false) assumption that everything is reducible to physical >>> substances and properties. >> >> I think that everything is reducible to physical substances and >> properties. And I think that all of physics is reducible to pure >> mathematics... > > You can't have it both ways. If physics was reducible to pure > mathematics, then physics could not be the 'ontological base level' of > reality and hence everything could not be expressed solely in terms of > physical substance and properties. Are you not begging a bit the question here?
> > Besides which, mathematics and physics are dealing with quite > different distinctions. It is a 'type error' it try to reduce or > identity one with the other. I don't see why.
> > Mathematics deals with logical properties, I guess you mean "mathematical properties". Since the filure of logicism, we know that math is not really related to logic in any way. It just happens that a big part of logic appears to be a branch of mathemetics, among many other branches. > physics deals with spatial > (geometric) properties. Although geometry is thought of as math, it > is actually a branch of physics,
Actually I do think so. but physics, with comp, has to be the science of what the observer can observe, and the observer is a mathematical object, and observation is a mathematical object too (with comp). > since in addition to pure logical > axioms, all geometry involves 'extra' assumptions or axioms which are > actually *physical* in nature (not purely mathematical) . Here I disagree (so I agree with your preceding post where you agree that we agree a lot but for not always for identical reasons). Arithmetic too need extra (non logical) axioms, and it is a matter of taste (eventually) to put them in the branch of physics or math. Bruno http://iridia.ulb.ac.be/~marchal/ |
| Re: Are First Person prime? |
George Levy |
26/11/07 11:22 |
Bruno
Yes I am particularizing things... But "the end justifies the means". I
am being positivist, trying to express these rules as a function of an
observer. In any case, once the specific example is worked out, we can
fall back on the general case.
Your feedback about "exist" not really being adequate to express truth
is well noted. Let me change the proposed rules to express truth as a
function of an axiomatic system A existing as data .... either in the
memory of M .... or as a axiomatic substrate for a simulated world
W..... Let's try the following:
In a world W simulated according to the axiomatic data system A, there
is a machine M, data p and data q such that
1) If M has access to p (possibly in its memory), then p exists in W.
(exist=being simulated in W according to A )
2) If M has access to p, then M has access to the access point to p.
3) If M has access to the information relating or linking p to q then
if M has access to p, it also has access to q.
Now we can make the statements reflexive ( I don't know if this is the
right word) by setting data p = Machine description M.
In a simulated world W following the axiomatic data system A there is a
machine M=p and data q such that
1) If M has access to M then M exists in W. (reflexivity?)
2) If M has access to M, then M has access to the access point to M.
(Infinite reflexivity? - description of consciousness?)
3) If M has information describing q as a consequence of M in
accordance with A, then
if M has access to M, it also has access to q. (This is a form of
Anthropic principle)
I am not sure if this is leading anywhere, but it's fun playing with
it. Maybe a computer program could be written to express these
staqtements.
George
|
| Re: Theory of Everything based on E8 by Garrett Lisi |
JohnM |
26/11/07 12:55 |
Listers, (Bruno, Torgny, et al.): some (lay) remarks from another mindset (maybe I completely miss your points - perhaps even my own ones<G>). I go with Bruno in a lack of clear understanding what "physical world" may be. It can be extended into entirely mathematical ideas beside the likable assumption of it being 'geometrical ' as well as geometry 'completely physical'. I don't see these terms agreed upon as crystal clearly (maybe my ignorance). * Then again "pure"(?) Math, the logical entirety, is in my views different from the "applied"(?) math of the diverse sciences, (please note the cap vs lower case distinction, as borrowed from the late mathematician Robert Rosen) the latter applying the former's results to quantities. (I don't want to digress here into my views about the restricted (topical) aspects of those sciences, omitting the rest of the totality that, however, may have an effect of those figments derived as 'scientific quantities' within their boundaries. It may come up in a separate (different) thread). To (I think) Torgny's remark "> > reality and hence everything could not be expressed solely in
terms of physical substance and properties.<<" I would add: also depends on a possible extension of the meaning 'physical'. * Then there is the reference to 'axioms'. These "true" postulates are formed AFTER a theory was thought through to maintain the validity of that theory. So I don't consider them "proof", rather as a consequence for the statement it is supposed to underlie. I believe these are Bruno's (supporting?) words: > Arithmetic too need extra (non logical) axioms, and it is a matter of taste (eventually) to put them in the branch of physics or math.<
* Please, excuse my 'out-of-context' remarks, I wanted to illustrate a different line of thoughts - also generated in a "human" mind. John M |
| Re: Theory of Everything based on E8 by Garrett Lisi |
Russell Standish |
26/11/07 14:33 |
Could we have a stop to HTML-only postings please! These are hard to read. On Mon, Nov 26, 2007 at 10:51:36AM +0100, Torgny Tholerus wrote: -- ---------------------------------------------------------------------------- A/Prof Russell Standish Phone 0425 253119 (mobile) Mathematics UNSW SYDNEY 2052 hpc...@hpcoders.com.au Australia http://www.hpcoders.com.au ----------------------------------------------------------------------------
|
| Re: Theory of Everything based on E8 by Garrett Lisi |
marc.geddes |
26/11/07 20:38 |
>When I talk about "pure mathematics" I mean that kind of mathematics you have in GameOfLife. There you have "gliders" that move in the GameOfLife-universe, and these gliders interact with eachother when they meet. These gliders you can see as physical objects. These physical objects are reducible to pure mathematics, they are the consequences of the rules behind GameOfLife.
--
Torgny
That kind of mathematics - models of cellular automata - is the
domain of the theory of computation. These are just that - models.
But there is no reason for thinking that the models or mathematical
rules are identical to the physical entities themselves just because
these rules/models can precisely predict/explain the behaviour of the
physical objects.
|
| Re: Theory of Everything based on E8 by Garrett Lisi |
marc.geddes |
26/11/07 20:47 |
On Nov 27, 3:54 am, Bruno Marchal <marc...@ulb.ac.be> wrote:
>
> > Besides which, mathematics and physics are dealing with quite
> > different distinctions. It is a 'type error' it try to reduce or
> > identity one with the other.
>
> I don't see why.
Physics deals with symmetries, forces and fields.
Mathematics deals with data types, relations and sets/categories.
The mathemtical entities are informational. The physical properties
are geometric. Geometric properties cannot be derived from
informational properties.
>
>
>
> > Mathematics deals with logical properties,
>
> I guess you mean "mathematical properties". Since the filure of
> logicism, we know that math is not really related to logic in any way.
> It just happens that a big part of logic appears to be a branch of
> mathemetics, among many other branches.
I would classify logic as part of applied math - logic is a
description of informational systems from the point of view of
observers inside time and space.
>
> > physics deals with spatial
> > (geometric) properties. Although geometry is thought of as math, it
> > is actually a branch of physics,
>
> Actually I do think so. but physics, with comp, has to be the science
> of what the observer can observe, and the observer is a mathematical
> object, and observation is a mathematical object too (with comp).
>
> > since in addition to pure logical
> > axioms, all geometry involves 'extra' assumptions or axioms which are
> > actually *physical* in nature (not purely mathematical) .
>
> Here I disagree (so I agree with your preceding post where you agree
> that we agree a lot but for not always for identical reasons).
> Arithmetic too need extra (non logical) axioms, and it is a matter of
> taste (eventually) to put them in the branch of physics or math.
>
> Bruno
>
I don't think it's a matter of taste. I think geoemtry is clearly
physics, arithmetic is clearly pure math. See above. Geometry is
about fields, arithmetic (in the most general sense) is about
categories/sets.
|
| Re: Theory of Everything based on E8 by Garrett Lisi |
Günther Greindl |
27/11/07 04:18 |
Dear Marc, > Physics deals with symmetries, forces and fields. > Mathematics deals with data types, relations and sets/categories. I'm no physicist, so please correct me but IMHO: Symmetries = relations Forces - could they not be seen as certain invariances, thus also relating to symmetries? Fields - the aggregate of forces on all spacetime "points" - do not see why this should not be mathematical relation? > The mathemtical entities are informational. The physical properties > are geometric. Geometric properties cannot be derived from > informational properties. Why not? Do you have a counterexample? Regards, Günther -- Günther Greindl Department of Philosophy of Science University of Vienna guenther...@univie.ac.at http://www.univie.ac.at/Wissenschaftstheorie/ Blog: http://dao.complexitystudies.org/ Site: http://www.complexitystudies.org
|
| Re: Theory of Everything based on E8 by Garrett Lisi |
Bruno Marchal |
27/11/07 06:16 |
Le 27-nov.-07, à 05:47, marc....@gmail.com a écrit : > Geometric properties cannot be derived from > informational properties.
I don't see why. Above all, this would make the computationalist wrong, or at least some step in the UDA wrong (but then which one?). I recall that there is an argument (UDA) showing that if comp is true, then not only geometry, but physics, has to be derived exclusively from numbers and from what numbers can prove (and know, and observe, and bet, ...) about themselves, that is from both extensional and intensional number theory. The UDA shows *why* physics *has to* be derived from numbers (assuming CT + "yes doctor"). The Lobian interview explains (or should explain, if you have not yet grasp the point) *how* to do that.
Bruno
http://iridia.ulb.ac.be/~marchal/ |
| Re: Are First Person prime? |
Bruno Marchal |
27/11/07 07:16 |
Le 26-nov.-07, à 20:22, George Levy a écrit : > Bruno Sure. And playing is the best way to learn, as "nature" knows since the beginning.
> Maybe a computer program could be written to express these staqtements.
I certainly encourage you to do so. Note that for the general modal theory, S4, or even the comp fist person S4Grz, programs already exists. By some aspect your attempt reminds me also of dynamic logic. This is modal logic applied to computations (and thus a bit away from "computability" which concerns my more "theological" global concern ;). You could googelize on "dynamic modal logic". I am not at all an expert on those logics, to be sure.
Bruno
http://iridia.ulb.ac.be/~marchal/
|
| OM = SIGMA1 links |
Bruno Marchal |
27/11/07 07:40 |
Hi Mirek, Brent, Barry, David, ... and all those who could be interested in the INTRO to Church thesis, I have to go, actually. Just to prepare yourself to what will follow, below are recent links in the list . It could be helpful to revise a bit, or to ask last questions. I will ASAP come back on Cantor's Diagonal, (one more post), and then I will send the key fundamental post where I will present a version of Church thesis, and explain how from just CT you can already derive what I will call the first fundamental theorem. This one says that ALL universal machine (if that exists) are insecure. It is needed to explain why Lobian machine, which are mainly just Universal machine knowing that they are universal, cannot not be above all "theological" machine. As you can guess, knowing that they are universal, will make them know that they are insecure. All the term here will be defined precisely. In case you find this theorem depressing, I suggest you read "The Wisdom of Insecurity" by Alan Watts (Pantheon Books, Inc. 1951). An amazingly "lobian" informal philosophical text. Here are the last posts I send: 1) Bijections 1 http://www.mail-archive.com/everything-lis...m/msg13962.html 2) Bijections 2 http://www.mail-archive.com/everything-lis...m/msg13986.html 3) Bijection 3 http://www.mail-archive.com/everything-lis...m/msg13991.html 4) Cantor's diagonal http://www.mail-archive.com/everything-lis...m/msg13996.html Don't hesitate to ask any question if something remains unclear, Bruno PS: I recall the combinators thread, which could help later (but please don't consult them now, unless you already love lambda calculus or the combinators).
The old (2005) combinators posts: http://www.mail-archive.com/everything-list@eskimo.com/msg05920.html http://www.mail-archive.com/everything-list@eskimo.com/msg05949.html http://www.mail-archive.com/everything-list@eskimo.com/msg05953.html http://www.mail-archive.com/everything-list@eskimo.com/msg05954.html http://www.mail-archive.com/everything-list@eskimo.com/msg05955.html http://www.mail-archive.com/everything-list@eskimo.com/msg05956.html http://www.mail-archive.com/everything-list@eskimo.com/msg05957.html http://www.mail-archive.com/everything-list@eskimo.com/msg05958.html http://www.mail-archive.com/everything-list@eskimo.com/msg05959.html http://www.mail-archive.com/everything-list@eskimo.com/msg05961.html http://iridia.ulb.ac.be/~marchal/
|
| Re: OM = SIGMA1 links |
Günther Greindl |
27/11/07 08:27 |
|
| Re: Theory of Everything based on E8 by Garrett Lisi |
marc.geddes |
27/11/07 20:40 |
On Nov 28, 1:18 am, Günther Greindl < guenther.grei...@gmail.com>
wrote:
> Dear Marc,
>
> > Physics deals with symmetries, forces and fields.
> > Mathematics deals with data types, relations and sets/categories.
>
> I'm no physicist, so please correct me but IMHO:
>
> Symmetries = relations
> Forces - could they not be seen as certain invariances, thus also
> relating to symmetries?
>
> Fields - the aggregate of forces on all spacetime "points" - do not see
> why this should not be mathematical relation?
>
> > The mathemtical entities are informational. The physical properties
> > are geometric. Geometric properties cannot be derived from
> > informational properties.
>
> Why not? Do you have a counterexample?
>
> Regards,
> Günther
>
Don't get me wrong. I don't doubt that all physical things can be
*described* by mathematics. But this alone does not establish that
physical things *are* mathematical. As I understand it, for the
examples you've given, what happens is that based on emprical
observation, certain primatives of geometry and symmetry are *attached
to* (connected with) mathematical relations, numbers etc which
successfully *describe/predict* these physical properties. But it
does not follow from this, that the mathematical relations/numbers
*are* the geometric properties/symmetrics.
In order to show that the physical properties *are* the mathematical
properties (and not just described by or connected to the physical
properties), it has to be shown how geometric/physical properties
emerge from/are logically derived from sets/categories/numbers alone. |
| Re: Theory of Everything based on E8 by Garrett Lisi |
marc.geddes |
27/11/07 20:48 |
On Nov 28, 3:16 am, Bruno Marchal < marc...@ulb.ac.be> wrote:
> Le 27-nov.-07, à 05:47, marc.ged...@gmail.com a écrit :
>
> > Geometric properties cannot be derived from
> > informational properties.
>
> I don't see why. Above all, this would make the computationalist wrong,
> or at least some step in the UDA wrong (but then which one?).
I'll find the flaw in UDA in due course ;)
> I recall that there is an argument (UDA) showing that if comp is true,
> then not only geometry, but physics, has to be derived exclusively from
> numbers and from what numbers can prove (and know, and observe, and
> bet, ...) about themselves, that is from both extensional and
> intensional number theory.
> The UDA shows *why* physics *has to* be derived from numbers (assuming
> CT + "yes doctor").
> The Lobian interview explains (or should explain, if you have not yet
> grasp the point) *how* to do that.
>
> Bruno
>
If the UDA is sound that would certainly refute what I'm claiming
yes. I want to see how physics (which as far I'm concerned *is*
geometry - at least I think pure physics=geometry) emerges *purely*
from theories of sets/numbers/categories.
I base my claims on ontological considerations (5 years of deep
thought about ontology), which lead me to strongly suspect the
irreducible property dualism between physical and mathematical
properties. Thus I'm highly skeptical of UDA but have yet to property
study it. Lacking resources to do proper study here at the
moment.... :-(
Time will tell. |
| Re: Theory of Everything based on E8 by Garrett Lisi |
Torgny Tholerus |
28/11/07 00:56 |
marc....@gmail.com
skrev:
You only need models of cellular automata. If you have a model and
rules for that model, then one event will follow after another event,
according to the rules. And after that event will follow another more
event, and so on unlimited. The events will follow after eachother
even if you will not have any implementation of this model. Any
physics is not needed. You don't need any geometric properties.
In this model you may have a person called Torgny writing a message on
a google group, and that event may be followed by a person called Marc
writing a reply to this message. And you don't need any implementation
of that model.
--
Torgny
|
| Re: Theory of Everything based on E8 by Garrett Lisi |
Quentin Anciaux |
28/11/07 02:04 |
Hi, Sure, but you can't be ultrafinitist and saying things like "And after that
event will follow another more event, and so on unlimited". Also why do you limit yourself to one computational model ? Turing Machine, Lambda calcul, cellular automata are all equivalents. Regards, Quentin Anciaux -- All those moments will be lost in time, like tears in the rain.
|
| Re: Theory of Everything based on E8 by Garrett Lisi |
Bruno Marchal |
28/11/07 05:51 |
Le 28-nov.-07, à 05:48, marc....@gmail.com a écrit : > > > > On Nov 28, 3:16 am, Bruno Marchal <marc...@ulb.ac.be> wrote: >> Le 27-nov.-07, à 05:47, marc.ged...@gmail.com a écrit : >> >>> Geometric properties cannot be derived from >>> informational properties. >> >> I don't see why. Above all, this would make the computationalist >> wrong, >> or at least some step in the UDA wrong (but then which one?). > > I'll find the flaw in UDA in due course ;) Thanks.
> >> I recall that there is an argument (UDA) showing that if comp is true, >> then not only geometry, but physics, has to be derived exclusively >> from >> numbers and from what numbers can prove (and know, and observe, and >> bet, ...) about themselves, that is from both extensional and >> intensional number theory. >> The UDA shows *why* physics *has to* be derived from numbers (assuming >> CT + "yes doctor"). >> The Lobian interview explains (or should explain, if you have not yet >> grasp the point) *how* to do that. >> >> Bruno >> > > If the UDA is sound that would certainly refute what I'm claiming > yes. > I want to see how physics (which as far I'm concerned *is* > geometry - at least I think pure physics=geometry) emerges *purely* > from theories of sets/numbers/categories.
OK. Note that UDA says only why, not how. "how" is given by the lobian interview, and gives only the "propositional physics" (as part of the propositional "theology"). > > I base my claims on ontological considerations (5 years of deep > thought about ontology), which lead me to strongly suspect the > irreducible property dualism between physical and mathematical > properties. Thus I'm highly skeptical of UDA but have yet to property > study it. Lacking resources to do proper study here at the > moment.... :-(
We are in the same boat ... Bruno http://iridia.ulb.ac.be/~marchal/ |
| Re: Theory of Everything based on E8 by Garrett Lisi |
Bruno Marchal |
28/11/07 05:57 |
Le 28-nov.-07, à 09:56, Torgny Tholerus a écrit : OK. Do you agree now that the "real Torgny", by which I mean you from your first person point of view, cannot known if it belongs to a state generated by automata 345 or automata 6756, or automata 6756690003121, or automata 65656700234676611084899 , and so one ... Do you agree we have to take into account this first person indeterminacy when making a first person prediction?
Bruno
http://iridia.ulb.ac.be/~marchal/
|
| Last post before the key post (was OM = SIGMA_1) 1 |
Bruno Marchal |
28/11/07 07:47 |
Le 27-nov.-07, à 17:27, Günther Greindl a écrit : > > Dear Bruno, > > thanks for your posts! I like them very much! > Looking forward to further stuff, Thanks for telling. In this post I recall Cantor's proof of the non enumerability of the set of infinite binary sequences. First with a drawing, and then with mathematical notation. The goal is to train you with the notations. Then I do the same with the almost exactly identical proof that the set of functions from N to N is also non enumerable.
I hope you have no more problem with the identification of the set of functions from N to 2 (where 2 = {0, 1}), with the set of infinite binary sequences. Please ask in case of any trouble with that. A function from N to any set A always gives rise to an infinite sequence of elements of A. Note that I will never use Cantor's result. I explain it just to illustrate the use of diagonalization, and to contrast it with the use in the next post which will illustrate a capital feature of all universal machine. *** 1) 2^N is not enumerable (proof by drawing) Cantor argument goes like this. It is a reductio ad absurdo: Suppose that the set of infinite binary sequence is enumerable. That means that there is a bijection between N and that set. Such a bijection will have a shape like: 0 -------- 000101101111010100111... 1 -------- 011111111010100110000... 2 -------- 100011001001000100110... 3 -------- 101010101001010100100... 4 -------- 100000000001001111010... 5 -------- 000111111111100000010... ... But then I can find an infinite binary sequence which *cannot* be in the image of that "bijection". Indeed here is one: 1 0 1 1 1 0 ... It is the complementary sequence build from the diagonal sequence. QED. *** 1bis) 2^N is not enumerable (proof without drawing). Suppose that the set of binary infinite sequences is enumerable. That means there is a bijection between N and that set. It means that for each natural number i, there is a corresponding sequence s_i, and that all infinite binary sequences belongs somewhere in the enumeration s_0 s_1 s_2 s_3 s_4 s_5 ... Each s_i is a function from N to 2. For example, if s_0 denote the first sequence in the "drawing" above, it means that s_0(0) = 0, s_0(1) = 0, s_0(2) = 0, s_0(3) = 1, s_0(4) = 0, s_0(5) = 1, s_0(6) =1, etc. That is, s_i(j) the jth number (0 or 1) in the sequence s_i. s_i(j) gives the matrix drawn above, for i and j natural numbers. Then the number s_i(i), for i natural numbers gives the diagonal sequence, and the sequence 1- s_i(i) gives the complementary of the diagonal, and that sequence cannot belongs to the list of the s_i. We can make explicit the contradiction. If the sequence 1- s_i(i) was in the list s_i, there would exist a number k such that s_k(x) = 1 - s_x(x). But then for x = k, s_k(k) = 1 - s_k(k). But s_k(k) has to be one or zero, and in the first case you get 1 = 1 - 1 = 0, and in the other case you get 0 = 1 - 0 = 1. Contradiction. Damn... I must already go. I suggest you work by yourself the very similar proof that N^N is not enumerable. Of course you can derive this immediately from what we have just seen, given that a function from N to 2, is a particular case of a function from N to N. But it is a good training in *diagonalisation* to find the direct proof.
I hope you can see that the set of all functions from N to N can be identify with the set NXNXNX... of sequences of natural numbers. I do quickly the "drawing proof", and let you write the more explicit proof (without drawing). Suppose there is such a bijection. It will look like: 0 -------- 23 456 76 67 ... 1 -------- 0 8 23 5 ... 2 -------- 67 10 10 123 ... 3 -------- 9 0 0 4 ... ... But then the sequence of numbers: 24 9 11 5 .... cannot be in that list. All right? See you tomorrow, Bruno http://iridia.ulb.ac.be/~marchal/ |
| Re: Theory of Everything based on E8 by Garrett Lisi |
Torgny Tholerus |
28/11/07 07:53 |
Bruno Marchal skrev: > > > Le 28-nov.-07, à 09:56, Torgny Tholerus a écrit : >
> You only need models of cellular automata. If you have a model > and rules for that model, then one event will follow after another > event, according to the rules. And after that event will follow > another more event, and so on unlimited. The events will follow > after eachother even if you will not have any implementation of > this model. Any physics is not needed. You don't need any > geometric properties. > > In this model you may have a person called Torgny writing a > message on a google group, and that event may be followed by a > person called Marc writing a reply to this message. And you don't > need any implementation of that model. > > > > OK. Do you agree now that the "real Torgny", by which I mean you from > your first person point of view, cannot known if it belongs to a state > generated by automata 345 or automata 6756, or automata 6756690003121, > or automata 65656700234676611084899 , and so one ... > Do you agree we have to take into account this first person > indeterminacy when making a first person prediction? I agree that the "real Torgny" belongs to exactly one of those automata, but I don't know which one. So I can not tell what will happen to the "real Torgny" in the future. I can not do any prediction. If we call the automata that the "real Torgny" belongs to, for automata X, then I can look at automata X from the outside, and I will then see that all that the "real Torgny" will do in the future is completely determined. There is no indeterminacy in automata X. -- Torgny
|
| Re: Last post before the key post (was OM = SIGMA_1) 1 |
Mirek Dobsicek |
28/11/07 08:32 |
Hi Bruno, I'm ready. Luckily, it is not long time ago, I've received my university degree in CS, so it was rather easy to follow :-) Sincerely, Mirek Bruno Marchal wrote: > > Le 27-nov.-07, à 17:27, Günther Greindl a écrit : > >> Dear Bruno, >> >> thanks for your posts! I like them very much! >> Looking forward to further stuff, > > > > Thanks for telling. > > > In this post I recall Cantor's proof of the non enumerability of the > set of infinite binary sequences. First with a drawing, and then with > mathematical notation. The goal is to train you with the notations. > Then I do the same with the almost exactly identical proof that the set > of functions from N to N is also non enumerable.
> ............ |
| Re: Theory of Everything based on E8 by Garrett Lisi |
marc.geddes |
28/11/07 20:46 |
On Nov 28, 9:56 pm, Torgny Tholerus <tor...@dsv.su.se> wrote:
>
> You only need models of cellular automata. If you have a model and
> rules for that model, then one event will follow after another event,
> according to the rules. And after that event will follow another more
> event, and so on unlimited. The events will follow after eachother even
> if you will not have any implementation of this model. Any physics is
> not needed. You don't need any geometric properties.
>
> In this model you may have a person called Torgny writing a message on a
> google group, and that event may be followed by a person called Marc
> writing a reply to this message. And you don't need any implementation
> of that model.
>
> --
> Torgny
A whole lot of unproven assumptions in there. For starters, we don't
even know that the physical world can be modelled solely in terms of
cellular automata at all. Digital physics just seems to be the latest
'trendy' thing, but actual evidence is thin on the ground.
Mathematics is much richer than just discrete math. Discrete math
deals only with finite collections, and as such is just a special case
of algebra. Algebraic relations extend beyond computational models.
Finally, the introduction of complex analysis, infinite sets and
category theory extends mathematics even further, beyond even
algebraic relations. So you see that cellular automata are only a
small part of mathematics as a whole. There is no reason for thinking
for that space is discrete and in fact physics as it stands deals in
continuous differential equations, not cellular automata.
Further, the essential point I was making is that an informational
model of something is not neccesserily the same as the thing itself.
An informational model of a person called Marc would capture only my
mind, not my body. The information has to be super-imposed upon the
physical, or embodied in the physical world. |
| Last post before the key post (was OM = SIGMA_1) 1bis |
Bruno Marchal |
29/11/07 07:47 |
Mirek, Le 28-nov.-07, à 17:32, Mirek Dobsicek a écrit : > > Hi Bruno, > > I'm ready. Luckily, it is not long time ago, I've received my > university > degree in CS, so it was rather easy to follow :-) > > Sincerely, > Mirek Thanks for telling me that you are ready. Now I feel a bit guilty because today and tomorrow I get unexpected work, and next week I am teaching again. I hope that those who have no university degree in CS have been able to follow the thread too.
I will try to resume the last exercise tomorrow, (one last post on Cantor's diagonal), and then, I will write, during next week, the key post, which will prove an absolutely fundamental theorem on the Universal Machines, a theorem without which UDA would be stuck in the sixth step, and without which the lobian interview would not make sense. The theorem says that ALL universal machines are insecure or imperfect. I guess some of you can already guess or produce the proof (in company of a general definition of "secure machine", 'course). Torgny,
You should be clearer about when you work *in* your ultrafinistic theory and when you work in its metatheory. If not, Quentin is right to ask you not to mention any sort of "infinite" of any kind. Most of the time, it is very hard to make sense of your approach, due to the lack of a clear distinction between the ultrafinistic theory and the informal metatheory you do refer to, very often. Note that without the movie graph (the 8th step of the UDA), comp remains coherent *only* through an explicitly physicalist version of ultrafinitism and an explicitly dualist theory of Mind (perhaps you should collaborate with Marc?). Mind would need matter (but then why, and what is it?), and the UDA would not go through because we would live in a unique and then very little universe. I guess everythingers would be skeptical at the start, here. Also the quantum facts are going in an opposite direction, imo. Actually, the movie-graph prevents such a move, I think. We can go back on this, later. To be sure I am open to critics there, I am not entirely satisfied with my presentation of the argument, and both George and Russell did succeed in making me thinking a lot more on that issue, or of the way to present it perhaps (more than I was expecting). Bruno http://iridia.ulb.ac.be/~marchal/ |
| Re: Theory of Everything based on E8 by Garrett Lisi |
Torgny Tholerus |
29/11/07 08:22 |
Quentin Anciaux skrev: > Hi, > > Le Wednesday 28 November 2007 09:56:17 Torgny Tholerus, vous avez écrit : > >>
>> You only need models of cellular automata. If you have a model and >> rules for that model, then one event will follow after another event, >> according to the rules. And after that event will follow another more >> event, and so on unlimited. The events will follow after eachother even >> if you will not have any implementation of this model. Any physics is >> not needed. You don't need any geometric properties. >> >>
> Sure, but you can't be ultrafinitist and saying things like "And after that > event will follow another more event, and so on unlimited". > There is a difference between "unlimited" and "infinite". "Unlimited" just says that it has no limit, but everything is still finite. If you add something to a finite set, then the new set will always be finite. It is not possible to create an infinite set.
So it is OK to use the word "unlimited". But it is not OK to use the word "infinite". Is this clear? Another important word is the word "all". You can talk about "all events". But in that case the number of events will be finite, and you can then talk about "the last event". But you can't deduce any contradiction from that, because that is forbidden by the type theory. And there will be more events after "the last event", because the number of events is "unlimited". As soon as you use the word "all", you will introduce a limit - all up to this limit. And you must then think of only doing conclusions that are legal according to type theory. So the best thing is to avoid the word "all" (and all synonyms of that word). -- Torgny |
| Re: Theory of Everything based on E8 by Garrett Lisi |
Quentin Anciaux |
29/11/07 08:38 |
Le Thursday 29 November 2007 17:22:59 Torgny Tholerus, vous avez écrit : > Quentin Anciaux skrev: > > Hi, > > > > Le Wednesday 28 November 2007 09:56:17 Torgny Tholerus, vous avez écrit : > >> You only need models of cellular automata. If you have a model and > >> rules for that model, then one event will follow after another event, > >> according to the rules. And after that event will follow another more > >> event, and so on unlimited. The events will follow after eachother even > >> if you will not have any implementation of this model. Any physics is > >> not needed. You don't need any geometric properties. > > > > Sure, but you can't be ultrafinitist and saying things like "And after > > that event will follow another more event, and so on unlimited". > > There is a difference between "unlimited" and "infinite". "Unlimited" > just says that it has no limit, but everything is still finite. If you > add something to a finite set, then the new set will always be finite. > It is not possible to create an infinite set. I'm sorry I don't get it... The set N as an infinite numbers of elements still every element in the set is finite. Maybe it is an english subtility that I'm not aware of... but in french I don't see a clear difference between "infini" and "illimité". > So it is OK to use the word "unlimited". But it is not OK to use the > word "infinite". Is this clear?
No, I don't see how a set which have not limit get a finite number of elements. > Another important word is the word "all". You can talk about "all > events". But in that case the number of events will be finite, and you > can then talk about "the last event". But you can't deduce any > contradiction from that, because that is forbidden by the type theory. > And there will be more events after "the last event", because the number > of events is "unlimited". If there are events after the last one, how can the last one be the last ? > As soon as you use the word "all", you will > introduce a limit - all up to this limit. And you must then think of > only doing conclusions that are legal according to type theory. o_O... could you explain what is type theory ? > So the best thing is to avoid the word "all" (and all synonyms of that > word). like everything ? Regards, Quentin Anciaux -- All those moments will be lost in time, like tears in the rain.
|
| Re: Theory of Everything based on E8 by Garrett Lisi |
Torgny Tholerus |
29/11/07 09:25 |
Quentin Anciaux skrev: > Le Thursday 29 November 2007 17:22:59 Torgny Tholerus, vous avez écrit : > >>
>> There is a difference between "unlimited" and "infinite". "Unlimited" >> just says that it has no limit, but everything is still finite. If you >> add something to a finite set, then the new set will always be finite. >> It is not possible to create an infinite set. >> > > I'm sorry I don't get it... The set N as an infinite numbers of elements still > every element in the set is finite. Maybe it is an english subtility that I'm > not aware of... but in french I don't see a clear difference between "infini" > and "illimité". > As soon as you talk about "the set N", then you are making a "closure" and making that set finite. The only possible way to talk about something without limit, such as natural numbers, is to give a "production rule", so that you can produce as many of that type of objects as you want. If you have a natural number n, then you can "produce" a new number n+1, that is the successor of n. > > >> So it is OK to use the word "unlimited". But it is not OK to use the >> word "infinite". Is this clear? >> > > No, I don't see how a set which have not limit get a finite number of > elements. > It is not possible for "a set" to have no limit. As soon as you construct "a set", then that set will always have a limit. Either you have to accept that the set N is finite, or you must stop talking about "the set N". It is enough to have a production rule for natural numbers. > >> Another important word is the word "all". You can talk about "all >> events". But in that case the number of events will be finite, and you >> can then talk about "the last event". But you can't deduce any >> contradiction from that, because that is forbidden by the type theory. >> And there will be more events after "the last event", because the number >> of events is "unlimited". >> > > If there are events after the last one, how can the last one be the last ? > The last event is the last event in the set of "all" events. But because you have a production rule for the events, it is always possible to produce new events after the last event. But these events do not belong to the set of "all" events. > >> As soon as you use the word "all", you will >> introduce a limit - all up to this limit. And you must then think of >> only doing conclusions that are legal according to type theory. >> > > o_O... could you explain what is type theory ? > Type theory is one of the solutions of Russel's paradox. You have a hierarchy of "types". Type theory says that the "all quantifiers" only can span objects of the same "type" (or lower types). When you create new objects, such that "the set of all sets that do not belong to themselves", then you will get an object of a higher "type", so that you can not say anything about if this set belongs to itself or not. The same thing with "the set of all sets". You can not say anything about if it belongs to itself. > >> So the best thing is to avoid the word "all" (and all synonyms of that >> word). >> > > like everything ? >
Yes... :-) -- Torgny
|
| Re: Theory of Everything based on E8 by Garrett Lisi |
Quentin Anciaux |
29/11/07 09:39 |
Le Thursday 29 November 2007 18:25:54 Torgny Tholerus, vous avez écrit : > Quentin Anciaux skrev: > > Le Thursday 29 November 2007 17:22:59 Torgny Tholerus, vous avez écrit : > >> There is a difference between "unlimited" and "infinite". "Unlimited" > >> just says that it has no limit, but everything is still finite. If you > >> add something to a finite set, then the new set will always be finite. > >> It is not possible to create an infinite set. > > > > I'm sorry I don't get it... The set N as an infinite numbers of elements > > still every element in the set is finite. Maybe it is an english > > subtility that I'm not aware of... but in french I don't see a clear > > difference between "infini" and "illimité". > > As soon as you talk about "the set N", then you are making a "closure" > and making that set finite. Ok then the set R is also finite ? > The only possible way to talk about > something without limit, such as natural numbers, is to give a > "production rule", so that you can produce as many of that type of > objects as you want. If you have a natural number n, then you can > "produce" a new number n+1, that is the successor of n. What is the production rules of the "no"set R ? > >> So it is OK to use the word "unlimited". But it is not OK to use the > >> word "infinite". Is this clear? > > > > No, I don't see how a set which have not limit get a finite number of > > elements. > > It is not possible for "a set" to have no limit. As soon as you > construct "a set", then that set will always have a limit. I don't get it. > Either you > have to accept that the set N is finite, or you must stop talking about > "the set N". It is enough to have a production rule for natural numbers. I don't accept and/or don't understand. > >> Another important word is the word "all". You can talk about "all > >> events". But in that case the number of events will be finite, and you > >> can then talk about "the last event". But you can't deduce any > >> contradiction from that, because that is forbidden by the type theory. > >> And there will be more events after "the last event", because the number > >> of events is "unlimited". > > > > If there are events after the last one, how can the last one be the last > > ? > > The last event is the last event in the set of "all" events. But > because you have a production rule for the events, it is always possible > to produce new events after the last event. But these events do not > belong to the set of "all" events. There exists no last element in the set N. > >> As soon as you use the word "all", you will > >> introduce a limit - all up to this limit. And you must then think of > >> only doing conclusions that are legal according to type theory. > > > > o_O... could you explain what is type theory ? > > Type theory is one of the solutions of Russel's paradox. You have a > hierarchy of "types". Type theory says that the "all quantifiers" only > can span objects of the same "type" (or lower types). When you create > new objects, such that "the set of all sets that do not belong to > themselves", then you will get an object of a higher "type", so that you > can not say anything about if this set belongs to itself or not. The > same thing with "the set of all sets". You can not say anything about > if it belongs to itself. > > >> So the best thing is to avoid the word "all" (and all synonyms of that > >> word). > > > > like everything ? > > Yes... :-) What you are saying seems like to me "So the best thing is to avoid words at all (and any languages)"... Regards, Quentin Anciaux -- All those moments will be lost in time, like tears in the rain.
|
| Re: Theory of Everything based on E8 by Garrett Lisi |
Torgny Tholerus |
29/11/07 09:52 |
Quentin Anciaux skrev: > Le Thursday 29 November 2007 18:25:54 Torgny Tholerus, vous avez écrit : > >>
>> As soon as you talk about "the set N", then you are making a "closure" >> and making that set finite. >> > > Ok then the set R is also finite ? > Yes. > >> The only possible way to talk about >> something without limit, such as natural numbers, is to give a >> "production rule", so that you can produce as many of that type of >> objects as you want. If you have a natural number n, then you can >> "produce" a new number n+1, that is the successor of n. >> > > What is the production rules of the "no"set R ? > How do you define "the set R"? -- Torgny
|
| Re: Theory of Everything based on E8 by Garrett Lisi |
Quentin Anciaux |
29/11/07 09:58 |
Le Thursday 29 November 2007 18:52:36 Torgny Tholerus, vous avez écrit : > Quentin Anciaux skrev: > > Le Thursday 29 November 2007 18:25:54 Torgny Tholerus, vous avez écrit : > >> As soon as you talk about "the set N", then you are making a "closure" > >> and making that set finite. > > > > Ok then the set R is also finite ? > > Yes. o_O > >> The only possible way to talk about > >> something without limit, such as natural numbers, is to give a > >> "production rule", so that you can produce as many of that type of > >> objects as you want. If you have a natural number n, then you can > >> "produce" a new number n+1, that is the successor of n. > > > > What is the production rules of the "no"set R ? > > How do you define "the set R"? http://en.wikipedia.org/wiki/Construction_of_real_numbers Choose your method... Regards, Quentin Anciaux -- All those moments will be lost in time, like tears in the rain.
|
| RE: Theory of Everything based on E8 by Garrett Lisi |
jessem |
29/11/07 10:08 |
---------------------------------------- > Date: Thu, 29 Nov 2007 18:25:54 +0100 > From: tor...@dsv.su.se > To: everyth...@googlegroups.com > Subject: Re: Theory of Everything based on E8 by Garrett Lisi > > > Quentin Anciaux skrev: >> Le Thursday 29 November 2007 17:22:59 Torgny Tholerus, vous avez écrit : >> >>> >>> There is a difference between "unlimited" and "infinite". "Unlimited" >>> just says that it has no limit, but everything is still finite. If you >>> add something to a finite set, then the new set will always be finite. >>> It is not possible to create an infinite set. >>> >> >> I'm sorry I don't get it... The set N as an infinite numbers of elements still >> every element in the set is finite. Maybe it is an english subtility that I'm >> not aware of... but in french I don't see a clear difference between "infini" >> and "illimité". >> > > As soon as you talk about "the set N", then you are making a "closure" > and making that set finite.
Why is that? How do you define the word "set"?
The only possible way to talk about > something without limit, such as natural numbers, is to give a > "production rule", so that you can produce as many of that type of > objects as you want. If you have a natural number n, then you can > "produce" a new number n+1, that is the successor of n.
Why can't I say "the set of all numbers which can be generated by that production ruler"? It almost makes sense to say a set is *nothing more* than a criterion for deciding whether something is a member of not, although you would need to refine this definition to deal with problems like Russell's "set of all sets that are not members of themselves" (which could be translated as the criterion, 'any criterion which does not match its own criterion'--I suppose the problem is that this criterion is not sufficiently well-defined to decide whether it matches its own criterion or not).
> >> >> >>> So it is OK to use the word "unlimited". But it is not OK to use the >>> word "infinite". Is this clear? >>> >> >> No, I don't see how a set which have not limit get a finite number of >> elements. >> > > It is not possible for "a set" to have no limit. As soon as you > construct "a set", then that set will always have a limit. Is there something intrinsic to your concept of the word "set" that makes this true? Is your concept of a set fundamentally different than my concept of well-defined criteria for deciding if any given object is a member or not?
Jesse |
| Re: Theory of Everything based on E8 by Garrett Lisi |
Torgny Tholerus |
29/11/07 10:28 |
Quentin Anciaux skrev: > Le Thursday 29 November 2007 18:52:36 Torgny Tholerus, vous avez écrit : > >> Quentin Anciaux skrev: >> >>
>>> What is the production rules of the "no"set R ? >>> >> How do you define "the set R"? >> > > http://en.wikipedia.org/wiki/Construction_of_real_numbers > > Choose your method... > The most important part of that definition is: 4. The order ? is /complete/ in the following sense: every non-empty subset of *R* bounded above <http://en.wikipedia.org/wiki/Upper_bound> has a least upper bound <http://en.wikipedia.org/wiki/Least_upper_bound>. This definition can be translated to: "If you have a production rule that produces rational numbers that are bounded above, then this production rule is producing a real number." This is the production rule for real numbers. -- Torgny
|
| Re: Theory of Everything based on E8 by Garrett Lisi |
Torgny Tholerus |
29/11/07 10:55 |
Jesse Mazer skrev: > > >> From: tor...@dsv.su.se >> >> >> As soon as you talk about "the set N", then you are making a "closure" >> and making that set finite. >> > > > Why is that? How do you define the word "set"? > > > The only possible way to talk about > >> something without limit, such as natural numbers, is to give a >> "production rule", so that you can produce as many of that type of >> objects as you want. If you have a natural number n, then you can >> "produce" a new number n+1, that is the successor of n. >> > > > Why can't I say "the set of all numbers which can be generated by that production ruler"?
As soon as you say "the set of ALL numbers", then you are forced to define the word ALL here. And for every definition, you are forced to introduce a "limit". It is not possible to define the word ALL without introducing a limit. (Or making an illegal circular definition...) > It almost makes sense to say a set is *nothing more* than a criterion for deciding whether something is a member of not, although you would need to refine this definition to deal with problems like Russell's "set of all sets that are not members of themselves" (which could be translated as the criterion, 'any criterion which does not match its own criterion'--I suppose the problem is that this criterion is not sufficiently well-defined to decide whether it matches its own criterion or not). > A "well-defined criterion" is the same as what I call a "production rule". So you can use that, as long as the criterion is well-defined. (What does the criterion, that decides if an object n is a natural number, look like?) > >> >> It is not possible for "a set" to have no limit. As soon as you >> construct "a set", then that set will always have a limit. >> > > > Is there something intrinsic to your concept of the word "set" that makes this true? Is your concept of a set fundamentally different than my concept of well-defined criteria for deciding if any given object is a member or not? > Yes, the definition of the word "all" is intrinsic in the concept of the word "set". -- Torgny
|
| RE: Theory of Everything based on E8 by Garrett Lisi |
jessem |
29/11/07 11:14 |
---------------------------------------- > Date: Thu, 29 Nov 2007 19:55:20 +0100 > From: tor...@dsv.su.se > To: everyth...@googlegroups.com > Subject: Re: Theory of Everything based on E8 by Garrett Lisi > >
> Jesse Mazer skrev: >> >> >>> From: tor...@dsv.su.se >>> >>> >>> As soon as you talk about "the set N", then you are making a "closure" >>> and making that set finite. >>> >> >> >> Why is that? How do you define the word "set"? >> >> >> The only possible way to talk about >> >>> something without limit, such as natural numbers, is to give a >>> "production rule", so that you can produce as many of that type of >>> objects as you want. If you have a natural number n, then you can >>> "produce" a new number n+1, that is the successor of n. >>> >> >> >> Why can't I say "the set of all numbers which can be generated by that production ruler"? > > As soon as you say "the set of ALL numbers", then you are forced to > define the word ALL here. And for every definition, you are forced to > introduce a "limit". It is not possible to define the word ALL without > introducing a limit. (Or making an illegal circular definition...) Why can't you say "If it can be generated by the production rule/fits the criterion, then it's a member of the set"? I haven't used the word "all" there, and I don't see any circularity either. > >> It almost makes sense to say a set is *nothing more* than a criterion for deciding whether something is a member of not, although you would need to refine this definition to deal with problems like Russell's "set of all sets that are not members of themselves" (which could be translated as the criterion, 'any criterion which does not match its own criterion'--I suppose the problem is that this criterion is not sufficiently well-defined to decide whether it matches its own criterion or not). >> > > A "well-defined criterion" is the same as what I call a "production > rule". So you can use that, as long as the criterion is well-defined. > > (What does the criterion, that decides if an object n is a natural > number, look like?) I would just define the criterion recursively by saying "1 is a natural number, and given a natural number n, n+1 is also a natural number".
Jesse |
| RE: Bijections (was OM = SIGMA1) |
jessem |
29/11/07 11:28 |
---------------------------------------- > Date: Tue, 20 Nov 2007 19:01:38 +0100 > From: tor...@dsv.su.se > To: everyth...@googlegroups.com > Subject: Re: Bijections (was OM = SIGMA1) > > > Bruno Marchal skrev: >> >> But infinite ordinals can be different, and still have the same >> cardinality. I have given examples: You can put an infinity of linear >> well founded order on the set N = {0, 1, 2, 3, ...}. >> The usual order give the ordinal omega = {0, 1, 2, 3, ...}. Now omega+1 >> is the set of all ordinal strictly lesser than omega+1, with the >> convention above. This gives {0, 1, 2, 3, ... omega} = {0, 1, 2, 3, 4, >> ....{0, 1, 2, 3, 4, ....}}. As an order, and thus as an ordinal, it is >> different than omega or N. But as a cardinal omega and omega+1 are >> identical, that means (by definition of cardinal) there is a bijection >> between omega and omega+1. Indeed, between {0, 1, 2, 3, ... omega} and >> {0, 1, 2, 3, ...}, you can build the bijection: >> >> 0--------omega >> 1--------0 >> 2--------1 >> 3--------2 >> ... >> n ------- n-1 >> ... >> >> All right? "-----" represents a rope. >> > An ultrafinitist comment: > > In the last line of this sequence you will have: > > ? --------- omega-1 > > But what will the "?" be? It can not be omega, because omega is not > included in N... > > -- > Torgny > There is no such ordinal as "omega-1" in conventional mathematics. Keep in mind that ordinals are always defined as sets of previous ordinals, with 0 usually defined as the empty set {}...So,
0 = {} 1 = {0} = {{}} 2 = {0, 1} = {{}, {{}}} 3 = {0, 1, 2} = {{}, {{}}, {{}, {{}}}} ...and so forth. In thes terms, the ordinal "omega" is the set of finite ordinals, or: omega = {0, 1, 2, 3, 4, ... } = too much trouble for me to write out in brackets How would the set "omega-1" be defined? It doesn't make sense unless you believe in a "last finite ordinal", which of course a non-ultrafinitist will not believe in. Jesse
|
| Re: Theory of Everything based on E8 by Garrett Lisi |
JohnM |
29/11/07 13:20 |
Marc, please, allow me to write in plain language - not using those fancy words of these threads. Some time ago when the discussion was in commonsensically more understandable vocabulary, I questioned something similar to Günther, as pertaining to "numbers" - the alleged generators of 'everything' (physical, quality, ideation, process, you name it). As Bruno then said: the positive integers do that - if applied in sufficiently long expressions. (please, Bruno, correct this to a bottom-low simplification) - I did not follow that and was promised some more explanatory text in "not so technical" language. The discussion over the past some weeks is even "more technical" for me. Is not the distinction relevant what I hold, that there are two kinds of 'number'-usage: the (pure, theoretical Math and the in sciences - (quantity related) - "applied math" - that uses the formalism (the results, even logics) of 'Math' to exercise 'math'? (Cap vs lower m) Geometry seems to be in between(????) and symmetry can be both, I think. I am no physicist AND no mathematician, (not even a logician), so I pretend to keep an objective eye on things in which I am not prejudiced by knowledge. (<G>). John M |
| Re: Theory of Everything based on E8 by Garrett Lisi |
Torgny Tholerus |
30/11/07 00:00 |
Jesse Mazer skrev: > > > >> Date: Thu, 29 Nov 2007 19:55:20 +0100 >> From: tor...@dsv.su.se >> >>
>> As soon as you say "the set of ALL numbers", then you are forced to >> define the word ALL here. And for every definition, you are forced to >> introduce a "limit". It is not possible to define the word ALL without >> introducing a limit. (Or making an illegal circular definition...) >> > > Why can't you say "If it can be generated by the production rule/fits the criterion, then it's a member of the set"? I haven't used the word "all" there, and I don't see any circularity either. What do you mean by a "well-defined criterion"? Is this a well-defined criterion? : The set R is defined by: (x belongs to R) if and only if (x does not belong to x). If it fits the criterion (x does not belong to x), then it's a member of the set R. Then we ask the question: "Is R a member of the set R?". How shall we use the criterion to answer that question? If we substitute R for x in the criterion, we will get: (R belongs to R) if and only if (R does not belong to R)... What is wrong with this? -- Torgny
|
| RE: Theory of Everything based on E8 by Garrett Lisi |
jessem |
30/11/07 01:11 |
---------------------------------------- > Date: Fri, 30 Nov 2007 09:00:17 +0100 > From: tor...@dsv.su.se > To: everyth...@googlegroups.com > Subject: Re: Theory of Everything based on E8 by Garrett Lisi > >
My instinct would be to say that a "well-defined" criterion is one that, given any mathematical object, will give you a clear answer as to whether the object fits the criterion or not. And obviously this one doesn't, because it's impossible to decide where R fits it or not! But I'm not sure if this is the right answer, since my notion of "well-defined criteria" is just supposed to be an alternate way of conceptualizing the notion of a set, and I don't actually know why "the set of all sets that are not members of themselves" is not considered to be a valid set in ZFC set theory. Jesse |
| Re: Theory of Everything based on E8 by Garrett Lisi |
Bruno Marchal |
30/11/07 06:02 |
Le 29-nov.-07, à 17:22, Torgny Tholerus a écrit : > > Quentin Anciaux skrev: >> Hi, >> >> Le Wednesday 28 November 2007 09:56:17 Torgny Tholerus, vous avez >> écrit : >> >>> >>> You only need models of cellular automata. If you have a model and >>> rules for that model, then one event will follow after another event, >>> according to the rules. And after that event will follow another >>> more >>> event, and so on unlimited. The events will follow after eachother >>> even >>> if you will not have any implementation of this model. Any physics >>> is >>> not needed. You don't need any geometric properties. >>> >>> >> Sure, but you can't be ultrafinitist and saying things like "And >> after that >> event will follow another more event, and so on unlimited". >> > > > There is a difference between "unlimited" and "infinite". "Unlimited" > just says that it has no limit, but everything is still finite. If you > add something to a finite set, then the new set will always be finite. > It is not possible to create an infinite set. Come on! Now you talk like a finitist, who accepts the idea of "potential infinity" (like Kronecker, Brouwer and the intuitionnist) and who rejects only the so called actual infinities, like ordinal and cardinal "numbers" (or sets).
At the ontic level, (or ontological, I mean the minimum we have to bet on at the third person pov), comp is mainly finitist. Judson Webb put comp (he calls it mechanism) in Finitism. But that is no more ultrafinitism. With finitism: every object of the "universe" is finite, but the universe itself is infinite (potentially or actually). With ultrafinitism, every object is finite AND the universe itself is finite too. Jesse wrote: How would the set "omega-1" be defined? It doesn't make sense unless you believe in a "last finite ordinal", which of course a non-ultrafinitist will not believe in. Actually this makes sense in what is called non standard model of arithmetic. In Peano Arithmetic you cannot define what is a finite number (the notion of finiteness is typically a second order notion). So it is consistent to add some "infinite numbers" in the model of PA. But you can prove in PA that any number, except zero, has a predecessor, so a non standard model of arithmetic can have infinite objects, but then also its predecessor, and all their successive predecessors. A non standard model will have the following order on it: 0 1 2 ... ... infinity-1 infinity infinity+1 .... (and possibly other so called Z-chain). Such order are NOT ordinals, note. All this is not important now, but I say this in passing. > My instinct would be to say that a "well-defined" criterion is one > that, given any mathematical object, will give you a clear answer as > to whether the object fits the criterion or not. And obviously this > one doesn't, because it's impossible to decide where R fits it or not! > But I'm not sure if this is the right answer, since my notion of > "well-defined criteria" is just supposed to be an alternate way of > conceptualizing the notion of a set, and I don't actually know why > "the set of all sets that are not members of themselves" is not > considered to be a valid set in ZFC set theory.
Frege and Cantor did indeed define or identify sets with their defining properties. This leads to the Russell's contradiction. (I think Frege has abandoned his work in despair after that). One solution (among many other one) to save Cantor's work from that paradox consists in formalizing set theory, which means using "belongness" as an undefined symbol obeying some axioms. Just two examples of an axiom of ZF (or its brother ZFC = ZF + axiom of choice): is the extensionality axiom: AxAyAz ((x b z <-> y b z) -> x = y) "b" is for "belongs". It says that two sets are equal if they have the same elements. AxEy(z included-in x -> z b y) with "z included-in x" is a macro for Ar(r b z -> r b x). This is the power set axiom, saying that the set of all subsets of some set is also a set).
Paradoxes a-la Russell are evacuated by restricting Jesse's "well-defined criteria" by 1) first order formula (in the set language, that is with "b" as unique relational symbols (+ equality) ... like the axioms just above. 2) but such first order formula have to be applied only to an already defined set. For example, you can defined the set of x such that x is in y and has such property P(x). With P defined by a set formula, and y an already defined set. Also, ZFC has the foundation axiom which forbids a set to belong to itself. In particular the informal collection of all sets which does not belongs to themselves is the universe itself, which cannot be a set (its power set would be bigger than the universe!). But there are version of ZF without foundation, or even with diverse versions of the negation of the foundation axiom, like Stephen Paul King appreciates so much. Actually, I will not use set theory at all, but I will mention often ZF and its many brothers (like ZFC, ZF+<some big cardinal>, ...) as example of very rich lobian machine. ZF can prove the complete (propositional) theology of a lesser rich (in provability power) machine like PA. Bruno PS BTW, on the FOR list Charles gives this interesting reference on Everett. It looks his son has made a movie on his father: http://www.newscientist.com/channel/opinion/mg19626311.800-interview- parallel-lives-can-never-touch.html http://iridia.ulb.ac.be/~marchal/
|
| N^N is not enumerable (post before key post bis) |
Bruno Marchal |
30/11/07 07:32 |
Hi, I recall the proof that N^N is not enumerable. I recall that N^N is the set of functions from N to N. Such functions associate a natural number to each natural number. Example: factorial = {(0, 1) (1, 1) (2, 2) (3, 6) (4, 24) (5 120) ...} of course the same information is provided by the sequence: 1, 1, 2, 6, 24, 120, .... which is an element of the infinite cartesian product NXNXNXNXNX .... So N^N is in bijection with NXNXNXNXNX ...., and giving that we are interested in cardinality, for all practical purpose we can identify both sets. Theorem; N^N is not enumerable
Proof 1 (by absurdo): If a bijection exists between N and N^N, then it will look like (cf the identification above): 0 ---------- 4 1 2 6 24 57 ... 1 ---------- 0 0 5 0 45 7 ... 2 ---------- 0 9 7 0 1 0 ... ... Well, at least in the mind of a God capable of seeing the whole infinite table. Again, for such a God, the diagonal is entirely well defined: 4 1 8 .... So we can consider the sequence of the diagonal elements, each added to one: 4+1 1+1 8+1 ..., that is 5 2 9 ... By construction this sequence is not in the list above, because it differs: from the 0th sequence, at the 0th "decimal" let us say, from the1st sequence, at the 1st "decimal", from the 2nd sequence, at the 2nd "decimal", from the 3rd sequence, at the 3nd "decimal" from the 4th sequence, at the 4th "decimal" from the 5th sequence, at the 5th "decimal" from the 6th sequence, at the 6th "decimal" from the 7th sequence, at the 7th "decimal" from the 8th sequence, at the 8th "decimal" ... from the n-th sequence, at the n-th "decimal" ... All right? This works for any tentative bijection proposed by any Gods. Now I give you the same proof, but with the traditional functional notation. This is for helping those who could have some notation problems. Please make you sure that you see I am giving the same proof, but with better notations: proof 2 (by absurdo)
If a bijection exists between N and N^N, then it means there is an enumeration of all functions from N to N, it looks like: f_0 f_1 f_2 f_3 f_4 f_5 f_6 ... All those f_i are well-defined (in Platonia) functions from N to N. So, the following function g is also well defined. By definition: g(n) = f_n(n) + 1 (this is the diagonal "+1" described above). Now g cannot be in the list f_0 f_1 f_2 ... Why? Because if g is in the list, it means there is a number k such that g = f_k (and thus for all n, g(n) = f_k(n)). But g(n) = f_k(n) + 1 by definition of g. Now we have, by applying g on its code k (cf g = f_k): g(k) = f_k(k) (by the assumption that g is in the list), and g(k) = f_k(k) + 1 (by definition of g) Thus (by Leibniz rule) f_k(k) = f_k(k)+1. But those are well defined numbers (cf: the f_k are functions from N to N), thus we can subtract f_k(k) on both sides, and this gives 0 = 1 Contradiction. Thus g cannot be in the list, and appears as a sheep without a rope. N^N has "more element than N", and is thus non enumerable. QED (OK?). NEXT: the key post. It should be even more simple, because, although we will be obliged to stay in Platonia, we will not invoke any God anymore. Instead we will invoke its quasi antipode, not the devil (!), but the humble finite earthly creature. The idea is to concentrate ourself on computable functions from N to N, instead of *all* functions from N to N. Computable? By who? Computability is an epistemic notion, it seems to involve a subject. Well, as you can guess, computable will mean "computable by that humble finite earthly creature" ... Now, what does that mean ...
Soon on a screen near you ;) ..., asap ... Good week-end, Bruno http://iridia.ulb.ac.be/~marchal/ |
| Re: Theory of Everything based on E8 by Garrett Lisi |
Torgny Tholerus |
30/11/07 11:00 |
Bruno Marchal skrev: > > > Le 29-nov.-07, à 17:22, Torgny Tholerus a écrit : >
> There is a difference between "unlimited" and "infinite". "Unlimited" > just says that it has no limit, but everything is still finite. If > you > add something to a finite set, then the new set will always be > finite. > It is not possible to create an infinite set. > > Come on! Now you talk like a finitist, who accepts the idea of > "potential infinity" (like Kronecker, Brouwer and the intuitionnist) > and who rejects only the so called actual infinities, like ordinal and > cardinal "numbers" (or sets). Yes, I am more like a finitist than an ultrafinitist in this respect. I accept that something can be without limit. But I don't want to use the word "potential infinity", because "infinity" is a meaningless word for me. > > At the ontic level, (or ontological, I mean the minimum we have to bet > on at the third person pov), comp is mainly finitist. Judson Webb put > comp (he calls it mechanism) in Finitism. But that is no more > ultrafinitism. With finitism: every object of the "universe" is > finite, but the universe itself is infinite (potentially or actually). > With ultrafinitism, every object is finite AND the universe itself is > finite too. Here I am an ultrafinitist. I believe that the universe is strictly finite. The space and time are discrete. And the space today have a limit. But the time might be without limit, that I don't know. > > Jesse wrote: > > My instinct would be to say that a "well-defined" criterion is one > that, given any mathematical object, will give you a clear answer > as to whether the object fits the criterion or not. And obviously > this one doesn't, because it's impossible to decide where R fits > it or not! But I'm not sure if this is the right answer, since my > notion of "well-defined criteria" is just supposed to be an > alternate way of conceptualizing the notion of a set, and I don't > actually know why "the set of all sets that are not members of > themselves" is not considered to be a valid set in ZFC set theory. > > Frege and Cantor did indeed define or identify sets with their > defining properties. This leads to the Russell's contradiction. (I > think Frege has abandoned his work in despair after that). > One solution (among many other one) to save Cantor's work from that > paradox consists in formalizing set theory, which means using > "belongness" as an undefined symbol obeying some axioms. Just two > examples of an axiom of ZF (or its brother ZFC = ZF + axiom of > choice): is the extensionality axiom: > AxAyAz ((x b z <-> y b z) -> x = y) "b" is for "belongs". It says that > two sets are equal if they have the same elements. > AxEy(z included-in x -> z b y) with "z included-in x" is a macro for > Ar(r b z -> r b x). This is the power set axiom, saying that the set > of all subsets of some set is also a set). For me "belongness" is not a problem, because everything is finite. For me the axiom of choice always is true, because you can always do a chioce in a finite world. > > Paradoxes a-la Russell are evacuated by restricting Jesse's > "well-defined criteria" by > 1) first order formula (in the set language, that is with "b" as > unique relational symbols (+ equality) ... like the axioms just above. > 2) but such first order formula have to be applied only to an already > defined set. This 2) rule is a very important restriction, and it is just this that my "type theory" is about. When you construct new things, those things can only be constructed from things that are already defined. So when you construct the set of all sets, then that new set will not be included in the new set. > For example, you can defined the set of x such that x is in y and has > such property P(x). With P defined by a set formula, and y an already > defined set. > > Also, ZFC has the foundation axiom which forbids a set to belong to > itself. This is a natural consequence of my type theory. When you construct a set, that set can never belong to itself, because that set is not defined before it is constructed. > In particular the informal collection of all sets which does not > belongs to themselves is the universe itself, which cannot be a set > (its power set would be bigger than the universe!). Yes, the set of all sets which does not belongs to themselves is the universe itself. But this is not a problem for me, because you can always extend the universe by creating new objects. So you can create the power set, and the power set will then be bigger than the universe. But this power set will not be part of the universe. -- Torgny
|
| Re: Theory of Everything based on E8 by Garrett Lisi |
Torgny Tholerus |
30/11/07 11:21 |
marc....@gmail.com skrev: > On Nov 28, 9:56 pm, Torgny Tholerus <tor...@dsv.su.se> wrote: > > >> You only need models of cellular automata. If you have a model and >> rules for that model, then one event will follow after another event, >> according to the rules. And after that event will follow another more >> event, and so on unlimited. The events will follow after eachother even >> if you will not have any implementation of this model. Any physics is >> not needed. You don't need any geometric properties. >> >> In this model you may have a person called Torgny writing a message on a >> google group, and that event may be followed by a person called Marc >> writing a reply to this message. And you don't need any implementation >> of that model. >> >>
> A whole lot of unproven assumptions in there. For starters, we don't > even know that the physical world can be modelled solely in terms of > cellular automata at all. Why can't our universe be modelled by a cellular automata? Our universe is very complicated, but why can't it be modelled by a very complicated automata? An automata where you have models for protons and electrons and photons and all other elementary particles, that obey the same laws as the particles in our universe? > Digital physics just seems to be the latest > 'trendy' thing, but actual evidence is thin on the ground. > Mathematics is much richer than just discrete math. Discrete math > deals only with finite collections, and as such is just a special case > of algebra. Isn't it enough with this special case? You can do a lot with finite collections. There is not any need for anything more. > Algebraic relations extend beyond computational models. > Finally, the introduction of complex analysis, infinite sets and > category theory extends mathematics even further, beyond even > algebraic relations. So you see that cellular automata are only a > small part of mathematics as a whole. There is no reason for thinking > for that space is discrete and in fact physics as it stands deals in > continuous differential equations, not cellular automata. > The reason why physics deals in continuous differential equations is that they are a very good approximation to a world where the distance between the space points and the time points are very, very small. And if you read a book in Quantum Field Theory, they often start from a discrete model, and then take the limit when the distances go to zero. > Further, the essential point I was making is that an informational > model of something is not neccesserily the same as the thing itself. > An informational model of a person called Marc would capture only my > mind, not my body. The information has to be super-imposed upon the > physical, or embodied in the physical world. >
If the model models every atom in your body, then that model will describe your body. That model will describe how the atoms in your body react with eachother, and they will describe all your actions. -- Torgny
|
| Re: Theory of Everything based on E8 by Garrett Lisi |
Quentin Anciaux |
01/12/07 03:52 |
And how this render the *set R* ]-infinity,infinity[ finite/limited or even the set N [0,infinity[ ? If as you say you have elements/events/... after the last element/event/... it is totally contradictory and meaning less to call it last... If I take it as a demonstration by absurd, then you've just demonstrated that there exists no last element/event/... How can you avoid this contradiction ? Regards, Quentin Anciaux -- All those moments will be lost in time, like tears in the rain.
|
| Re: Theory of Everything based on E8 by Garrett Lisi |
Bruno Marchal |
03/12/07 02:55 |
Le 30-nov.-07, à 20:00, Torgny Tholerus a écrit : > Here I am an ultrafinitist. I believe that the universe is strictly > finite. The space and time are discrete. And the space today have a > limit. But the time might be without limit, that I don't know. Then you are physicalist before being ultrafinitist. Now ultrafinitism implies comp (OK?, 'course comp does not imply ultrafinitism!) But I have already argued that comp implies the falsity of physicalism (UDA), so? BTW, you often quote wiki or other standard definition of math concept. But few are justifiable in the ultrafinitist realm, so many of your statements seems contradictory to me. Bruno http://iridia.ulb.ac.be/~marchal/ |
| Re: Theory of Everything based on E8 by Garrett Lisi |
Bruno Marchal |
03/12/07 06:52 |
Le 30-nov.-07, à 20:21, Torgny Tholerus a écrit : > Why can't our universe be modelled by a cellular automata? By UDA, this is just a priori impossible.
What *is* still possible, is that you can "modelize" the emergence of the appearance of a universe by modelling, with a cellular automata, a couple "observer + a quantum cellular automata", or by modelling all possible observers through universal dovetaling. Normally the appearance of the quantum will be generated as well. If you are a machine, the physical universe (the sharable third person pov) is not describable in term of working machine. Unless you are the whole universe yourself, which I doubt. If, you are the whole universe, and if you (the universe) exist, and if comp is false and ultrafinitism true, then you are right. But then, about the mind body problem, you are reintroducing the material bullet making even impossible to really addressed the question. Imo: it would be a regression. This comment assumes a good understanding of the seven first steps of the UDA. > Our universe > is very complicated, but why can't it be modelled by a very complicated > automata? Because, by assuming comp, the (physical) universe has to emerge non locally from an infinity of infinite computations.
> An automata where you have models for protons and electrons > and photons and all other elementary particles, that obey the same laws > as the particles in our universe?
Of course I talk here on exact emulation. FAPP, you can simulate electron and photon. About "reality", I am not even sure there are photons and electrons, we have non local (in our local histories/branches) wavy interacting fields. Note that Newton's law, taken seriously enough, are also not turing emulable, like almost everything in naïve math.
Bruno
http://iridia.ulb.ac.be/~marchal/ |
| Re: Cantor's Diagonal |
David Nyman |
03/12/07 07:56 |
On Nov 20, 4:40 pm, Bruno Marchal < marc...@ulb.ac.be> wrote:
> Conclusion: 2^N, the set of infinite binary sequences, is not
> enumerable.
>
> All right?
OK. I have to try to catch up now, because I've had to be away longer
than I expected, but I'm clear on this diagonal argument.
David
> Hi,
>
> David, are you still there? This is a key post, with respect to the
> "Church Thesis" thread.
>
> So let us see that indeed there is no bijection between N and 2^N =
> 2X2X2X2X2X2X... = {0,1}X{0,1}X{0,1}X{0,1}X... = the set of infinite
> binary sequences.
>
> Suppose that there is a bijection between N and the set of infinite
> binary sequences. Well, it will look like that, where again "----"
> represents the "ropes":
>
> 0 -------------- (1, 0, 0, 1, 1, 1, 0 ...
> 1 -------------- (0, 0, 0, 1, 1, 0, 1 ...
> 2 --------------- (0, 1, 0, 1, 0, 1, 1, ...
> 3 --------------- (1, 1, 1, 1, 1, 1, 1, ...
> 4 --------------- (0, 0, 1, 0, 0, 1, 1, ...
> 5 ----------------(0, 0, 0, 1, 1, 0, 1, ...
> ...
>
> My "sheep" are the natural numbers, and my neighbor's sheep are the
> infinite binary sequences (the function from N to 2, the elements of
> the infinite cartesian product 2X2X2X2X2X2X... ).
> My flock of sheep is the *set* of natural numbers, and my neighbor's
> flock of sheep is the *set* of all infinite binary sequences.
>
> Now, if this:
>
> 0 -------------- (1, 0, 0, 1, 1, 1, 0 ...
> 1 -------------- (0, 0, 0, 1, 1, 0, 1 ...
> 2 --------------- (0, 1, 0, 1, 0, 1, 1, ...
> 3 --------------- (1, 1, 1, 1, 1, 1, 1, ...
> 4 --------------- (0, 0, 1, 0, 0, 1, 1, ...
> 5 ----------------(0, 0, 0, 1, 1, 0, 1, ...
> ...
>
> is really a bijection, it means that all the numbers 1 and 0 appearing
> on the right are well determined (perhaps in Platonia, or in God's
> mind, ...).
>
> But then the diagonal sequence, going from the left up to right down,
> and build from the list of binary sequences above:
>
> 1 0 0 1 0 0 ...
>
> is also completely well determined (in Platonia or in the mind of a
> God).
>
> But then the complementary sequence (with the 0 and 1 permuted) is also
> well defined, in Platonia or in the mind of God(s)
>
> 0 1 1 0 1 1 ...
>
> But this infinite sequence cannot be in the list, above. The "God" in
> question has to ackonwledge that.
> The complementary sequence is clearly different
> -from the 0th sequence (1, 0, 0, 1, 1, 1, 0 ..., because it differs at
> the first (better the 0th) entry.
> -from the 1th sequence (0, 0, 0, 1, 1, 0, 1 ... because it differs at
> the second (better the 1th) entry.
> -from the 2th sequence (0, 0, 0, 1, 1, 0, 1 ... because it differs at
> the third (better the 2th) entry.
> and so one.
> So, we see that as far as we consider the bijection above well
> determined (by God, for example), then we can say to that God that the
> bijection misses one of the neighbor sheep, indeed the "sheep"
> constituted by the infinite binary sequence complementary to the
> diagonal sequence cannot be in the list, and that sequence is also well
> determined (given that the whole table is).
>
> But this means that this bijection fails. Now the reasoning did not
> depend at all on the choice of any particular bijection-candidate. Any
> conceivable bijection will lead to a well determined infinite table of
> binary numbers. And this will determine the diagonal sequence and then
> the complementary diagonal sequence, and this one cannot be in the
> list, because it contradicts all sequences in the list when they cross
> the diagonal (do the drawing on paper).
>
> Conclusion: 2^N, the set of infinite binary sequences, is not
> enumerable.
>
> All right?
>
> Next I will do again that proof, but with notations instead of
> drawing, and I will show more explicitly how the contradiction arise.
>
> Exercice-training: show similarly that N^N, the set of functions from N
> to N, is not enumerable.
>
> Bruno
>
> http://iridia.ulb.ac.be/~marchal/ |
| Re: Cantor's Diagonal |
Bruno Marchal |
04/12/07 06:10 |
Le 03-déc.-07, à 16:56, David Nyman a écrit : > > On Nov 20, 4:40 pm, Bruno Marchal <marc...@ulb.ac.be> wrote: > >> Conclusion: 2^N, the set of infinite binary sequences, is not >> enumerable. >> >> All right? > > OK. I have to try to catch up now, because I've had to be away longer > than I expected, but I'm clear on this diagonal argument. OK. I hope you have done the exercise below, which consists to show that N^N is also not enumerable. It is the same reasoning, of course. (Of course it is also a consequence of the non enumerability of 2^N, because 2^N is included in N^N, but in the exercise I was asking for a direct diagonal argument). You can look at the solution: http://www.mail-archive.com/everything-list@eskimo.com/msg14063.html I will send the "key post" asap. It is written, but I am currently hesitating to add more explanations, but sometimes when I do that, I introduce just more confusion. I guess I will send some part so that you have something to think about (of 'course I don't doubt you have many things to think about ...). Bruno =================================================================== http://iridia.ulb.ac.be/~marchal/ |
| Key Post 1, toward Church Thesis and Lobian machine |
Bruno Marchal |
04/12/07 06:55 |
Hi David, Mirek, Tom, Barry and All, In the preceding post, I gave you an informal proof in naïve set theory that the set of functions from N to N was not enumerable. Note: the preceding post = http://www.mail-archive.com/everything-list@eskimo.com/msg14063.html It is not in my intention at all to convince you that Cantor's result is true. A problem which occurs is the many invocation of Gods. Cantor himself will hide paradoxes and also will discuss with theologians on those problems.
Today we know Cantor's theorem is provable in reasonably sound accounts of sets, like Zermelo Fraenkel Set Theory, ZF. But it is not my purpose to dwelve into set theory. My purpose is computability theory. What happens if we try to restrict ourselves to *computable* functions
from N to N, instead of *all* functions from N to N. In this case we would not be obliged to refer to Gods, those who were able to "see" an arbitrary and infinite long sequence of numbers. But we get somehow the anti-problem. How to define what is a computable function? We can hardly be satisfied by an anti-solution like: Anti-definition: A function from N to N is computable if it can be computed without any invocation to any Gods. So we have to find a more positive definition, and invoke *finite and humble* creature instead. Humble, because we have to be sure not attributing to them any magical and hard to define qualities like cleverness, intelligence, intuition or any god-like predicate. So here is an informal "definition" of what is an (intuitively) computable function from N to N.
"Definition": a function f is computable if there is a finite set of instructions such that a complete asshole can, on each n compute in a finite time the result f(n).
OK. "complete asshole" is probably a bit popular, and I will use the adverb "unambiguously" instead.
Also, what are the instructions? Whatever they are, they have to be described in some language, which has to be unambiguous (understandable by that humble finite creature).
So here is a better "definition" of what is a computable function from N to N. A function f from N to N is said (intuitively) computable if there is a language L in which it is possible to describe unambiguously through a finite expression in L how, for any n to compute f(n) in a finite time. The finite expression are intended to express the instructions. A language is the given of a finite alphabet A, and a subset L of unambiguous expressions. If A is a finite or enumerable alphabet, then I will denote by A°, the set of finite expressions written with the alphabet A. I recall that If A is finite or enumerable, then A° is enumerable too. Indeed, you can put an alphabetical order on all sequences of length 1, and then of length 2, etc. Exemple: A = {S, K, (, ) } Ordering the finite expressions, using the order "K < S < ( < )" on the alphabet:
finite expressions= K, S, (, ), [length one] KK, KS, K(, K), SK, SS, S(, S), (K, (S, ((, (), )K, )S, )(, )), [length two] KKK , ... [length 3] KKKK, ... [length 4] Note that ")))KK))S())" is a finite expression on the alphabet. It is does not refer to a combinator, which are associated only to well-formed expressions, like, if you remember, (K(SK)), or ((S(KS))K), making a subset of the set of all (finite) expressions. Now, two fundamental definitions:
Universal Language: A language L is universal if all computable functions (from N to N) can be described in L. Universal Machine: A machine M is universal if M understands L, and so M can actually compute the value f(n) of any computable function from its description in the universal language L and the input n. (Note that such a universal *machine*, should be describable itself by an expression in the universal language. We will come back on this later). Now the question is: Is there an universal language? Is there a universal machine? Is that an obvious question? Definitions like above are not proof of existence. Traditionnaly here I do sometimes present a proof, by diagonalization, that there are no universal machine, and ask the student to find possible errors. Here I will NOT proceed like that and proceed directly instead. For this I will first consider the problem of the cardinality of the set of computable functions, and then provide more definitions. The cardinality of the set of computable functions. Well, if L is a language, it has a finite alphabet A. Then, the subset of its unambiguous expressions (for the instruction) is a subset of the set of all its finite expressions, which we have seen to be enumerable. So the set of computable functions from N to N is enumerable. By Cantor, the set of functions from N to N is not enumerable: thus there are drastically more uncomputable functions than computable functions. Definition: Perfect Universal Machine (or Language): I will say that a universal machine (or language) is perfect, or secure, if the machine computes (or the language defines) only computable functions from N to N. By universality such a machine computes all computable functions from N to N. By security or perfection, such machine computes thus all and only all computable functions from N to N. By the definition of a function from N to N (in a preceding post), this means that such a universal secure machine will, on any unambiguous expression representing its instructions, output a value f(n) for any n, after a finite time. So a perfect universal machine, when following its instruction, never crash, by which I mean going in some loop or infinite task, well, that is, loosing the contact with the user or some neighborhood. Fundamental Theorem 1 : there are no *secure* universal machine. All universal machine, if there is ever one, have to be imperfect. Proof (reductio ad absurdo)
Suppose there is a secure universal machine M. The set of expressions it can compute provide a secure universal language L. That set is not only enumerable (given that it is a subset of an enumerable set) but above all, it can be enumerated effectively (by the "ashole"). Indeed the set of all unambiguous expressions, which, by perfection, describes computable functions from N to N, is enumerable by the method described above. Assuming the existence of such an universal secure language, we can effectively enumerate all the computable functions from N to N: f_1 f_2 f_3 f_4 f_5 f_6 f_7 f_8 ... where f_i represents the function computed by the i-th expression. But then the function g defined on each n by g(n) = f_n (n) + 1 is a computable function from N to N. To compute g on 777, for example, search the 777th expression in L by the method above, and apply it on 777, then add 1. But g cannot be computed by the secure universal machine M: indeed, if M compute g, it means there is an expression in L explaining how to compute g, and thus there is a k such that g = f_k, but then, applying g on its number: g(k) = f_k (k) on one hand, (presence of f_k in the list by universality) and g(k) = f_k(k) + 1 on the other hand, by definition of g. That is f_(k) = f_k (k) + 1 But the machine never crashes and computes all and only all functions from N to N, so f_k(k) is a number, so I can substract it on both sides, giving 0 = 1. Contradiction. QED. Remarks: the proof looks like Cantor proof that N^N is not enumerable. Here we know at the start that N^N, once restricted to computable functions, is enumerable. So if we know that a machine is a secure machine, we know that (by definition) all its expression defines functions from N to N, so we know that the machine cannot be universal. If we believe that a machine is a universal, then we can deduce two things: it has to compute some other beats Bs. Then, no set of instruction can for sure distinguish the ... instructions defining functions from N to N and the Bs. Proof: from such a set of instruction you could securized the universal machine, and get a perfecr one, which is impossible by the 1.
Any machine M is left with a choice: security or universality, if that exists. But never both! But does a universal language, and its corresponding universal (and thus insecure) machine, really exist? We still don't know that! That means: does it exist a machine computing all computable functions? Equivalently: does it exist a universal language in which all computable function can be described by an expression. Of course now we know that if such a universal machine exists, it will be insecure and will compute, not just all computable functions from N to N, but also other sort of beast. So, as you know, Church did *define* a computable function by a function computable by a lambda expression, in its conversion calculus.
Kleene tought it could refute Church's pretension of universality, by *diagonalising* against the set of lambda expressions. Let us look what happens. We don't have to define the lambda expressions or any other candidate to universality to make the following reasoning, which is the same as above, yet in a different context again! We are now in the context of someone presenting us with a well specified language L and pretending it is universal: it does compute all computable functions from N to N, he says! Now, the unambiguous expressions U_i of the language L * can* again be effectively enumerated: U_1 U_2 U_3 U_4 U_5 ... Each expression like that denotes now either a computable function from N to N, or as we have to expect something else. And we have to expect they are no computable means to distinguish which U_i represents functions from N to N, and which represents the other beast. Now, the only thing which can happen when following instruction and not giving a number as output, in this case, is that the process of computation run for ever. Due to this, we have an argument for the consistency of Church thesis. Indeed, the diagonalisation above, where now the f_i are the *partial computable functions*, meaning they are from N to N, OR from a subset of N to N, does no more lead to a contradiction. That f_i is the perhaps total perhaps not, function computed by the U_i: f_1 f_2 f_3 f_4 f_5 f_6 f_7 f_8 Yes the function g defined, (but no more necessarily on each n) by g(n) = f_n (n) + 1
can be represented by an expression in the language. So there is a k such that g = f_k indeed. And thus g(k) = f_k(k), and g(k) = f_k(k)+1 Right. So what happens if the universal machine compute g(k) ? Well, in the computer jargon, the machine crash. The poor humble creature go in loop, or in some infinite task without ever any thought for the user until this one reboot the system. But thanks to that crashing, Church thesis remains consistent. I have to go ... Please ask questions. By experience I know this can be confusing: we do always the same diagonalisation, and get different results. Of course the premises are different. I let you think a bit, before resuming and proceeding trough. Please, ask questions if anything remains unclear. The motivation is that a Lobian Machine will be mainly a sort of enlightened universal machine, i.e. a universal machine knowing that she is universal. "knowing" in a very weak and technical sense which will be made precise in due course. Bruno http://iridia.ulb.ac.be/~marchal/ |
| Re: Key Post 1, toward Church Thesis and Lobian machine |
Mirek Dobsicek |
05/12/07 14:08 |
Hi Bruno, thank you for your post. I read it a couple of times in order to more or less grasp it, but it worth it. I have some questions... > Suppose there is a secure universal machine M. The set of expressions > it can compute provide a secure universal language L. That set is not > only enumerable (given that it is a subset of an enumerable set) but > above all, it can be enumerated effectively (by the "ashole").
What do you mean here by effectively? I understand it as you just want to emphasize that the set is really enumerable. > So, as you know, Church did *define* a computable function by a > function computable by a lambda expression, in its conversion calculus.
Why do you introduce here the term 'lambda expression'? I'm asking this because in the sequel you work just with 'a well specified language which is promised to be universal' and you prove that such a promise is not ruled out. I do not see how you reached the conlusion: "But thanks to that crashing, *Church thesis remains consistent*. I would just say "An existence of a universal language is not ruled out". > Each expression like that denotes now either a computable function from > N to N, or as we have to expect something else. And we have to expect > they are no computable means to distinguish which U_i represents > functions from N to N, and which represents the other beast.
Can I say that the other beasts are only and only infinite loops? I assume that the machine cannot destroy itself, so it either stops after computing a computable function or enters some silly loop. > Indeed, the diagonalisation above, where now the f_i are the *partial > computable functions*, meaning they are from N to N, OR from a subset > of N to N, does no more lead to a contradiction.
I have a problem with this paragraph, could you please write more on this? I understand to partial computable functions as to functions which are not *defined* for every possible input (total functions are defined for all inputs, in my limited understanding). Do you say in that paragraph that beasts are only and only these partial functions? Huh, now what do I mean by *defined* ... maybe I should say 'which are not computable for every possible input'. I am really lost here... And my last question, consider the profound function f such that f(n) = 1 if there is a sequence of n consecutive fives in the decimal expansion of PI, and f(n) = 0 otherwise
Is this an example of a partial computable function? Or is this function as such already considered as un-computable function? With best regards, Mirek
|
| Re: Key Post 1, toward Church Thesis and Lobian machine |
Russell Standish |
05/12/07 14:23 |
On Wed, Dec 05, 2007 at 11:08:34PM +0100, Mirek Dobsicek wrote: > > > Each expression like that denotes now either a computable function from > > N to N, or as we have to expect something else. And we have to expect > > they are no computable means to distinguish which U_i represents > > functions from N to N, and which represents the other beast. > > Can I say that the other beasts are only and only infinite loops? I > assume that the machine cannot destroy itself, so it either stops after > computing a computable function or enters some silly loop. > Not unless the total number of states was finite. In a Turing machine case, the tape being infinite and readable/writeable allows the machine to compute forever without entering a loop. For instance, a program outputting the digits of Pi onto the tape computes forever and never enters a loop (since Pi is irrational, periodic sequences of digits are ruled out). --
---------------------------------------------------------------------------- A/Prof Russell Standish Phone 0425 253119 (mobile) Mathematics UNSW SYDNEY 2052 hpc...@hpcoders.com.au Australia http://www.hpcoders.com.au ----------------------------------------------------------------------------
|
| Re: Key Post 1, toward Church Thesis and Lobian machine |
Russell Standish |
05/12/07 15:01 |
On Tue, Dec 04, 2007 at 03:55:50PM +0100, Bruno Marchal wrote: > > Hi David, Mirek, Tom, Barry and All, >
... > > The cardinality of the set of computable functions. >
Thanks for this post. I was in the position of trying to explain your work to someone (actually a son of my mother's cousin) at a dinner party a couple of weeks ago, and having explained Cantor's diagonalisation proof of the uncountability of the reals, I got to the point about computable functions being countable and got stuck. I just had to say "well its true, but I can't quite recall the proof!". Your exposition is eminently dinner-party standard, although I might use a different word than "asshole"! Cheers -- ---------------------------------------------------------------------------- A/Prof Russell Standish Phone 0425 253119 (mobile) Mathematics UNSW SYDNEY 2052 hpc...@hpcoders.com.au Australia http://www.hpcoders.com.au ----------------------------------------------------------------------------
|
| Re: Key Post 1, toward Church Thesis and Lobian machine |
Bruno Marchal |
06/12/07 06:37 |
Thanks Russell. About the use of "asshole" I am afraid it is more popular, or vulgar, than I thought. You are very kind to tell me. Should I use "dumb" instead? The idea consists in not attributing anything like intuition, intelligence, cleverness, etc. for the followers of unambiguous instructions. It will be clearer when I will give example of "universal language", but I really do not want to do that to much quickly, because a main point here is that we can get "rigorously" many incompleteness and unsolvability results on formal language and machine without using any specific machine or formal theory. This is a key for learning to separate the different level of reasoning we will have to do. I hope you did not disturb too much the appetite of your mother's cousin's son :) Cheers, Bruno Le 06-déc.-07, à 00:01, Russell Standish a écrit :
http://iridia.ulb.ac.be/~marchal/ |
| Re: Key Post 1, toward Church Thesis and Lobian machine |
Bruno Marchal |
06/12/07 07:42 |
Hello Mirek, Le 05-déc.-07, à 23:08, Mirek Dobsicek a écrit :
> thank you for your post. I read it a couple of times in order to more > or > less grasp it, but it worth it. I have some questions... > > >> Suppose there is a secure universal machine M. The set of expressions >> it can compute provide a secure universal language L. That set is not >> only enumerable (given that it is a subset of an enumerable set) but >> above all, it can be enumerated effectively (by the "ashole"). > > What do you mean here by effectively? I understand it as you just want > to emphasize that the set is really enumerable.
I guess I will come back on this in the next post. But the point is that a set can be enumerable, yet NOT effectively enumerable. Effectively enumerable means enumerated by following unambiguous instructions in some language (universal or not). A dumb idiotic can do that.
To sum up a bit: Recall that by a function from A to B, I always mean a function defined for all elements of A having value in B. In particular a function from N to N is defined on all natural number. To emphasize on this I will say sometimes "TOTAL function". Then I will say "strictly partial function for a function from S to N, when S is a proper subset of N. Cantor's diagonal on N^N shows that the set of functions from N to N, i.e. N^N, is not enumerable (uncountable, ... mathematician have a lot of synonyms). So there is no bijection from N to N^N. Let us define N^N-comp as the set of computable functions in the sense described in the "key post". It is the set of total computable functions, or better described (but less usual) it is the total functions which are computable. Then N^N-comp is enumerable (indeed their corresponding set-of-instructions is a subset of all expressions in the language, which is already enumerable). Diagonalisation on N^N-comp shows that, although N^N-comp is enumerable, it is *not effectively enumerable*. The bijection between N^N-comp and N exist, yes, but is not a computable function. If it was, then the diagonal function g (with g(n) = f_n(n) +1) would be computable, in the list, and then 0 = 1, as I have try to explain. So it is really the bijection sending n on f_n which is not computable, and which makes g not computable. We can still believe there is a universal language in which we can define all total computable function, but then the language has to define more than the total computable one. In that case we get a language L which defines a more general class of "computable functions", the one which are no more defined on all inputs. In that case the diagonal function g can be defined in the language, but then such function has to be undefined on its "godel number k", that is, the number k such that g = f_k. And the key point is that no set of instructions at all can help the universal machine to distinguish in all case a code of a total function from a code of a strictly partial function. For if that was possible, then we could "securise" the universal machine making it computing only total functions, and then the diagonal will strike again and prove 0 = 1. Note this discovery: If A is enumerable, and if B is included in A, then B is enumerable. But if A is effectively enumerable (meaning really that the ass..., er I mean the dumb one can enumerate A), it DOES NOT follow that any subset of A is effectively enumerable. There is a bijection between the set of code (instructions, expressions) of total computable function and N, but what the second diagonalization shows, is that such a bijection is not a computable bijection. The set of code of total function is an enumerable, but not effectively enumerable subset of the set of code of the partial (strict and not strict) functions. > > >> So, as you know, Church did *define* a computable function by a >> function computable by a lambda expression, in its conversion >> calculus. > > Why do you introduce here the term 'lambda expression'? I'm asking this > because in the sequel you work just with 'a well specified language > which is promised to be universal' and you prove that such a promise is > not ruled out. > I do not see how you reached the conlusion: > "But thanks to that crashing, *Church thesis remains consistent*. I > would just say "An existence of a universal language is not ruled out". OK. But Church thesis says that Lambda is universal, and so a weaker form of Church thesis (the one which asserts the existence of a universal language) remains consistent. OK. > > >> Each expression like that denotes now either a computable function >> from >> N to N, or as we have to expect something else. And we have to expect >> they are no computable means to distinguish which U_i represents >> functions from N to N, and which represents the other beast. > > Can I say that the other beasts are only and only infinite loops? I > assume that the machine cannot destroy itself, so it either stops after > computing a computable function or enters some silly loop.
Despite universal machine are finite entity (like us with comp), we let them write on wall, papers, or magnetic tape, .... So as Russell said, the machine could enter in a infinite and never repeated task. Of course if the machine ask for more memory, and if its sadic user just say no, well the machine will have problem. later I will insist heavily that universal machine are finite object. There are important reason of not putting the "infinite tape" in the machine. The tape of a Turing machine (for those who met her) is not part of the machine.
> > >> Indeed, the diagonalisation above, where now the f_i are the *partial >> computable functions*, meaning they are from N to N, OR from a subset >> of N to N, does no more lead to a contradiction. > > I have a problem with this paragraph, could you please write more on > this? I understand to partial computable functions as to functions > which > are not *defined* for every possible input (total functions are defined > for all inputs, in my limited understanding). That is correct. Note that such a definition makes the total function a particular case of the partial function. Total function: they are defined for any n. Partial function: they are defined for any n, or not. It is possible that for some n, they will be undefined.
> Do you say in that > paragraph that beasts are only and only these partial functions? Yes.
> > Huh, now what do I mean by *defined* ... maybe I should say 'which are > not computable for every possible input'. I am really lost here...
OK I see the ambiguity in the wording. Let me try another wording: a function is a total function (from N to N) if it is defined on all N. Example, the factorial function, the successor function, etc. Then a function will be said to be a partial function from N to N, if it is a (total) function from a subset of N to N. Now, just remember that N is a particular case of subset of N (a non *proper*) subset, it is said. So a total function from N to N is a particular case of partial function. That N is a subset of N, follows from the fact that x belongs to N implies that x belongs to N. (A is included in B iff (by definition) x b A -> x b B (b = belongs). Did this help? Don't worry. Some problem can occur because the word function is not used with the same meaning in the world (flemish and french talker already use different definition). We can decide to be clear on that given the international character of the mailing list. I am sure people use different version, and that is why I have always (re) define the notion, but I should have insisted that such a word has just no standard definition). Even mathematicians does not always use the term function with always the same meaning, when they go from one branch of math to another, like when going from arithmetic to analysis. But all serious mathematician are aware of that and usually redefine such terms before using them. > > > And my last question, consider the profound function > f such that f(n) = 1 if there is a sequence of n consecutive fives in > the decimal expansion of PI, and f(n) = 0 otherwise > > Is this an example of a partial computable function? Or is this > function > as such already considered as un-computable function?
This is "certainly" an example of a total function from N to N. I put "certainly" in quote because I am using the principle of the excluded middle. If you agree that for each n, either there are n consecutive fives in the decimal expansion of PI, or they are not n consecutive fives in the decimal expansion of PI, then you agree that, on, your function f is equal to 1 or to 0. So the function, for a platonist, is perfectly well define on each n. It is a total function from N to N. Is such a function computable? That is an open problem. What is clear, is that nobody can compute it today, but this really proves nothing. Open question. Note that for the bijection between the set of codes of total function and N, we are in a different situation: such a function has been shown (in the key post btw) to be not computable by any universal machine, as defined even just informally like I have done. With Church thesis (the half of comp) such a function will never been computable (by finite humble creature).
Hope this helps a bit. I have to go, so I will not *add* spelling mistakes, this time (g). Don't hesitate to ask any question. Take the needed time for digesting. Best, Bruno http://iridia.ulb.ac.be/~marchal/ |
| Re: Key Post 1, toward Church Thesis and Lobian machine |
Russell Standish |
06/12/07 15:22 |
On Thu, Dec 06, 2007 at 03:37:10PM +0100, Bruno Marchal wrote: > > Thanks Russell. > > About the use of "asshole" I am afraid it is more popular, or vulgar, > than I thought. You are very kind to tell me. > Should I use "dumb" instead? The idea consists in not attributing No "dumb" is the wrong word. Dumb people can of course be quite creative (literally dumb means not able to speak, but derogatively it also means "stupid", which is a slight on dumb people). I would have gone with something like "mindless robot" - although this has problems of implying artificiality. Perhaps "mindless servant" - you want to get across the idea of following orders to the letter without question. Doesn't "trou de cul" have a similar meaning to "arsehole" in French? I suspect the closer translation might be "connard" though. --
---------------------------------------------------------------------------- A/Prof Russell Standish Phone 0425 253119 (mobile) Mathematics UNSW SYDNEY 2052 hpc...@hpcoders.com.au Australia http://www.hpcoders.com.au ----------------------------------------------------------------------------
|
| Re: Key Post 1, toward Church Thesis and Lobian machine |
Bruno Marchal |
07/12/07 03:31 |
Le 07-déc.-07, à 00:22, Russell Standish a écrit : > > On Thu, Dec 06, 2007 at 03:37:10PM +0100, Bruno Marchal wrote: >> >> Thanks Russell. >> >> About the use of "asshole" I am afraid it is more popular, or vulgar, >> than I thought. You are very kind to tell me. >> Should I use "dumb" instead? The idea consists in not attributing > > No "dumb" is the wrong word. Dumb people can of course be quite > creative (literally dumb means not able to speak, but derogatively it > also means "stupid", which is a slight on dumb people). All right. I knew only the derogative meaning. > > I would have gone with something like "mindless robot" - although this > has problems of implying artificiality.
Ah Ah .... I was expecting someone asking if the "asshole" I did mention (sorry), was not just the one we are used to call "machine". As you say, this implies in general some artificiality, but not necessarily so. After all the doctrine of Mechanism (related to Descartes) is the thesis that we (actually only animals for Descartes) are machine. And this does not mean we are artificial. BTW note that the term "artificial" is artificial" itself, and so is natural, and thus artificial, and thus natural, and thus artificial .... But yes, in the informal ,pregodelian sense of machine, the term "machine" is not so bad (and obviously more polite than the one I used).
> Perhaps "mindless servant" - > you want to get across the idea of following orders to the letter > without question.
Mindless servant is rather good too, with the informal sense. But saying that a function is computable if it can be computed without mind can be considered as a definition as negative as saying that a computation can be done without invoking Gods. Term like "mind", "God" are a bit problematic when used at the start of the enquiry. > > Doesn't "trou de cul" have a similar meaning to "arsehole" in French? Yes, literally. But "trou de cul" is a bit old fashioned. > I suspect the closer translation might be "connard" though.
Or the simpler "con", which is used a lot, but it is not only vulgar, but also disrespectful for woman (like most insults in french). Best, Bruno http://iridia.ulb.ac.be/~marchal/
|
| Re: Key Post 1, toward Church Thesis and Lobian machine |
Bruno Marchal |
11/12/07 07:21 |
Hi Brent, Mirek, David, .... From what you told me, I think you have no problem with Cantor 's diagonal.
Are you ok with the key post, that is with the two supplementary uses of the diagonal in the enumerable context? Let me sum up, please consult the preceding posts for details. 1) Cantor:
If f_1 f_2 f_3 f_4 .... represents the image of a candidate bijection from N to N^N, then the diagonal function g, defined by g(n) = f_n(n) + 1 gives a counter-example, that is, a function from N to N, which is not in the image of the candidate bijection. This works for any candidate bijection, so we can conclude that there are no bijections between N and N^N. 2) We restrict the set of all functions from N to N. Now we consider that the functions have to be computable, and this means that there exist a language in which we can define those functions.
Lemma (= preliminary proposition): the set of things definable in a language L is enumerable. (All right?) Is there a universal language in which we can define all (total) computable function from N to N? a) Theorem: the answer to that question is NO, if it is asked that all expressions (= well formed instructions for one variable function, say) in the language define all AND ONLY ALL computable functions. Proof: if it was the case that all the expressions (for function of one variable) defines all total computable function from N to N, then, by the lemma, the set of such functions are not only enumerable, but can be enumerated mechanically, computably, etc. .... b) Theorem: if the answer is yes, then a universal machine cannot be securized by any machine. .... Oops: I'm interrupted. I let you try to finish this post. I come back on it friday, or next week. Please don't hesitate to ask me any question, or to make any remark, including meta-remarks, jokes, or whatever. It would help me to have an idea if most of you get the point or if most of you did not get it ... It is simple, but admittedly not so simple, sure. Bruno
http://iridia.ulb.ac.be/~marchal/
|
| Re: Key Post 1, toward Church Thesis and Lobian machine |
Mirek Dobsicek |
12/12/07 10:58 |
Hi Bruno, > From what you told me, I think you have no problem with Cantor 's > diagonal. Yep, no problem. > Are you ok with the key post, that is with the two supplementary uses > of the diagonal in the enumerable context? 95% grasped, and for the rest I'm lacking time to do a sufficient amount of scribes in order to get it completely. But nothing fundamental ... Now, I'm very busy with finishing my phd thesis (study and simulations of a certain two-qubit procedure, a sort of benchmark). I guess, I'll be fine at the beginning of the New Year. Sincerely, Mirek |
| Re: Key Post 1, toward Church Thesis and Lobian machine |
Barry Brent |
12/12/07 16:27 |
|
| Re: Key Post 1, toward Church Thesis and Lobian machine |
Bruno Marchal |
14/12/07 06:18 |
Hi Barry and Mirek, (and Brent, David, ....). Thanks for telling,
New year is good for me. As you know I am a bit of a platonist so time has no real meaning for me. I told you that this year I'm teaching Church thesis at my Saturday Course on computer science for a large (not necessarily mathematician) public, and I am much slow there than here (we have not yet begin the bijection!). I will take the time to make all things clear, even for those without any knowledge in math, but of course it would help if they dare to ask questions. The key post is certainly not perfect, and will evolve thanks to you and my students. In the meantime I will provide a little summary (which I have already try to make, but it goes repeatedly in the trash because it is too long). I intent also to give some sample of "universal system". By a universal system, I mean the whole complex of a universal machine, its formal universal language, the set of its instructions, codes, etc. Let me give you already an example or two.
The Shepherdson Sturgis coffee-bar formal definition of computability. (A variant by Cutland). Here is a job offer in an (infinite) coffee bar in Platonia. (Infinite, just for making things a bit simpler.)
The basic instructions are the following 3 types + 1. a. - Please add a coffee cup on table 17 (say) b.- Please put on table 24 as many coffee cups than there are coffee cups on table 42 (say) c. - Please make sure there is no more coffee cups on table 56 The last instruction is a bit more difficult. To do the job you need minimal ability to read a language in which the preceding instructions can be described. Also, to economize paper (yes in Platonia Forest are protected too!), the instruction a. - Please add a coffee cup on table 17 (say) is written S(17)
b.- Please put on table 24 as many coffee cups than there are coffee cups on table 42 (say) is written T(42, 24) c. - Please make sure there is no more coffee cups on table 56 is written Z(56) (Z is for zero cup of coffee) For the last instruction you have to know that a job is the given of an ordered, even numbered instructions. For example, a typical Job is 1) Z(1) 2) S(1) 3) S(2) Here the job consists in making sure there are no more coffee cups on table one. Then to add a cup of coffe on table one, and then add a cup of coffee on table 2. Now here is the last instruction: d. - if the number of coffee cups on table 14 (say) is equal to the number of coffee cups on table 45 (say) then proceed from the instruction 5 (say) described in your job. In case there are no instruction numbered 5, stop (the job will be said to be completed); in case the number of coffee cups on table 14 is not equal to the number of coffee cups on table 45, then proceed from the next instruction. It is written: J(14, 45, 5).
DEFINITION: A function f from N to N is said to be Shepherdson Sturgis coffee bar computable, if there is a job (a list of numbered instructions) such that when putting n cups of coffee on table one, then, after the job is completed there is f(n) cups of coffee on table one. Similarly, a function h from NXN to N is said to be Shepherdson Sturgis coffee bar computable if there is a job such that, after having put n cups of coffee on table one and m cups of coffee on table two, then, after the job is completed there is h(n,m) cups of coffee on table one.
I have to go, so I give some Exercise for the week-end (I provide solution monday) 1) find a short job "crashing" the coffee bar computer. Such a job will never be completed. 2) find a job which computes addition (which is of course a function from NXN to N) 3) using the preceding job, find a job which computes multiplication. 4) is the following proposition plausible: a function from N to N is intuitively computable if and only if it can be computed by some coffee bar job. 5) describe informally the coffee-bar language, and, choosing an order on its alphabet, write the first 7 jobs in the lexicographical order. The alphabet contains all symbols needed in the jobs, including commas, parentheses, etc. + some grammatical rules making clear that Z(23) is a good instruction, but 23(Z) is not, ... Bruno
Le 13-déc.-07, à 01:27, Barry Brent a écrit : http://iridia.ulb.ac.be/~marchal/ |
| Re: Cantor's Diagonal |
Dan Grubbs |
16/12/07 01:49 |
Hi Folks,
I joined this list a while ago but I haven't really kept up. Anyway, I
saw the reference to Cantor's Diagonal and thought perhaps someone
could help me.
Consider the set of positive integers: {1,2,3,...}, but rather than
write them in this standard notation we'll use what I'll call 'prime
notation'. Here, the number m may be written as
m = n1,n2,n3,n4,...
where ni is the number of times the i'th prime number is a factor of m.
Thus:
1 = 0,0,0,0,0,...
2 = 1,0,0,0,0,...
3 = 0,1,0,0,0,...
4 = 2,0,0,0,0,...
5 = 0,0,1,0,0,...
...
28 = 2,0,0,1,0,0,0,...
...
If we now apply the diagonal method to this ordered set, we get the
number:
D = 1,1,1,1,1,...
Has this just shown that the set of positive integers is not
denumerable?
I can see that one may complain that D is clearly infinite and
therefore should not be in the set, but consider the following...
Let's take the original set and reorder it by exchanging the places of
the i'th prime number with that of the number in the i'th position.
(i.e. First switch the number 2 with the number 1 to move it to the
first position. Then switch 3 with the number -- now 1 -- in the 2nd
position. Then 5 with the 1 which is now in the 3rd position. Etc...)
Because we are just trading the positions of the numbers, all the same
numbers will be in the set afterwards.
The set is now:
2 = 1,0,0,0,0,...
3 = 0,1,0,0,0,...
5 = 0,0,1,0,0,...
7 = 0,0,0,1,0,...
11= 0,0,0,0,1,...
...
Now instead of adding 1 to each 'digit' of the diagonal, subtract 1.
This will ensure that the diagonal number is different from each of the
numbers in the set. Thus,
D = 0,0,0,0,...
But this is the number 1 which we know was in the set to begin with.
What happened to it?
I would suggest that the diagonal method does not find a number which
is different from all the members of a set, but rather finds a number
which is infinitely far out in the ordered set.
If anyone can find where I've gone wrong, please let me know.
Dan Grubbs
|
| Re: Cantor's Diagonal |
Barry Brent |
16/12/07 12:03 |
Hi. Bruno could do this better, but I like the practice. I guess you're trying to demonstrate that the form of Cantor's argument is invalid, by displaying an example in which it produces an absurd result. Start with a set S you want to show is not enumerable. (ie, there is no one-one onto map from Z^+ to S). The form of the diagonalization argument is as follows: give me any, repeat, any, particular candidate for an enumeration of S. This should be a map from Z^+ into S. (If it isn't such a map, it isn't an enumeration.) I will show you an element D of S that your candidate enumeration omits. (That is, I will show you that your candidate is not onto.) Hence, S is not denumerable. In your first attempt, your D is not an element of your S (= Z^+). So your first attempt doesn't fit the form of the diagonalization argument on this account. More fundamentally, it also fails to fit the form of Cantor's argument because you haven't tried to debunk *any* candidate enumeration, but a particular one. In your second attempt, if I understand you, you start with a map from the primes (all of them!) and then (your work suggests, but I think you'd need more details--what's the image of 4, for example?) from the rest of Z^+, into S = Z^+ again. This example doesn't invalidate Cantor's argument either. Again, you debunk a particular candidate enumeration, not any and all candidate enumerations. So you don't arrive at the absurdity you seem to be after, even if you fill in the details I mentioned. Barry Dr. Barry Brent barry...@earthlink.net http://home.earthlink.net/~barryb0/ |
| Re: Cantor's Diagonal |
Russell Standish |
16/12/07 15:10 |
On Sun, Dec 16, 2007 at 04:49:34AM -0500, Daniel Grubbs wrote: Cantor's argument only works by finding a number that satisfies the criteria for inclusion in the list, yet is nowhere to be found in the list. In your first case, the number (1,1,1,1...) is not a natural number, since it is infinite. In the second case, (0,0,0,...) is a natural number, but is also on the list (at infinity). Therefore Cantor's argument doesn't work in these cases. Cheers -- ---------------------------------------------------------------------------- A/Prof Russell Standish Phone 0425 253119 (mobile) Mathematics UNSW SYDNEY 2052 hpc...@hpcoders.com.au Australia http://www.hpcoders.com.au ----------------------------------------------------------------------------
|
| Re: Cantor's Diagonal |
Dan Grubbs |
16/12/07 19:09 |
Hi Barry,
Let me see if I am clear about Cantor's method. Given a set S, and
some enumeration of that set (i.e., a no one-one onto map from Z^+ to
S) we can use the diagonalization method to find an D which is a valid
element of S but is different from any particular indexed element in
the enumeration.
Cantor's argument then goes on to say (and here is where I disagree
with it) that therefore D is not included in the enumeration and the
enumeration is incomplete.
I, on the other hand, would posit that the enumeration may include
elements whose index is not well defined. For example, 1 or 4 in my
second example. They were in the enumeration prior to its being
shuffled, and always had a definite position during the process. They
must still be in there, with definite positions, despite the fact that
their indices are now infinite and ill-defined.
If the diagonalization process does not produce the proffered result in
this case (i,e., it does not prove that the element D is not included
in the enumeration) then it does not prove it in any case. The number
D found with this method may actually be in the enumeration, but with
an ill-defined index.
Dan
|
| Re: Cantor's Diagonal |
Barry Brent |
16/12/07 22:27 |
Hi Dan, Let me take your statements a few at a time. >> Let me see if I am clear about Cantor's method. Given a set S, >> and some enumeration of that set (i.e., a no one-one onto map from >> Z^+ to S) we can use the diagonalization method to find an D >> which is a valid element of S but is different from any particular >> indexed element in the enumeration. Not "given a set S and *some* enumeration..." but "given a set S and *any candidate for an enumeration*..." The idea is to show that no candidate for an enumeration succeeds, hence the set S has no enumeration, hence the set S is not enumerable. >> we can use the diagonalization method to find an D which is a >> valid element of S but is different from any particular indexed >> element in the enumeration. Yes. >> Cantor's argument then goes on to say (and here is where I >> disagree with it) that therefore D is not included in the >> enumeration and the enumeration is incomplete. Yes, except for your disagreement with it.... One step of Cantor's argument is to show, for a given candidate enumeration of S, that some D in S is different from every single enumerated member of S. If you can't show this for some particular candidate enumeration, you haven't done a Cantor diagonalization argument that encompasses the candidate enumeration. To do a successful Cantor diagonalization argument, you must show that some D in S is different from every single enumerated member of S, for each candidate enumeration of S. The value of D will typically depend on the candidate. >> I, on the other hand, would posit that the enumeration may include >> elements whose index is not well defined. For example, 1 or 4 in >> my second example. They were in the enumeration prior to its >> being shuffled, and always had a definite position during the >> process. They must still be in there, with definite positions, >> despite the fact that their indices are now infinite and ill-defined. I don't need to parse this shuffling enumeration to say that what you're doing is not an instance of Cantor's argument. Any instance of Cantor's argument will debunk, not just one candidate shuffling enumeration, to take your example, but each and every candidate enumeration. And you haven't even debunked the shuffling enumeration. In fact you're arguing that it's valid (it seems to me.) >> If the diagonalization process does not produce the proffered >> result in this case (i,e., it does not prove that the element D is >> not included in the enumeration) then it does not prove it in any >> case. The number D found with this method may actually be in the >> enumeration, but with an ill-defined index. A defender of the validity of the form of Cantor's argument doesn't *want* to use it to prove that Z^+ has no enumeration. To do so would be to fall into the paradox you're trying to construct. So it's perfectly fine if the diagonalization process does not produce "the proffered result" in this case. Barry On Dec 16, 2007, at 9:09 PM, Daniel Grubbs wrote: > Hi Barry, > > Let me see if I am clear about Cantor's method. Given a set S, > and some enumeration of that set (i.e., a no one-one onto map from > Z^+ to S) we can use the diagonalization method to find an D which > is a valid element of S but is different from any particular > indexed element in the enumeration. > > Cantor's argument then goes on to say (and here is where I disagree > with it) that therefore D is not included in the enumeration and > the enumeration is incomplete. > > I, on the other hand, would posit that the enumeration may include > elements whose index is not well defined. For example, 1 or 4 in > my second example. They were in the enumeration prior to its being > shuffled, and always had a definite position during the process. > They must still be in there, with definite positions, despite the > fact that their indices are now infinite and ill-defined. > > If the diagonalization process does not produce the proffered > result in this case (i,e., it does not prove that the element D is > not included in the enumeration) then it does not prove it in any > case. The number D found with this method may actually be in the > enumeration, but with an ill-defined index. > > Dan Dr. Barry Brent barry...@earthlink.net http://home.earthlink.net/~barryb0/ |
| Re: Cantor's Diagonal |
Bruno Marchal |
17/12/07 06:22 |
Hi Daniel, I agree with Barry, but apaprently you have still a problem, so I comment your posts. Le 16-déc.-07, à 10:49, Daniel Grubbs a écrit :
> Hi Folks, > > I joined this list a while ago but I haven't really kept up. Anyway, > I saw the reference to Cantor's Diagonal and thought perhaps someone > could help me. > > Consider the set of positive integers: {1,2,3,...}, but rather than > write them in this standard notation we'll use what I'll call 'prime > notation'. OK. That is often used to compare the purely additive and the purely multiplicative structures of the natural numbers. Of course the number 0 is banished in the multiplicative structure (so you have to handle 0 manualy, but that does not change the enumeration question, ok). > Here, the number m may be written as >> m = n1,n2,n3,n4,... >
>> 1 = 0,0,0,0,0,... >> 2 = 1,0,0,0,0,... >> 3 = 0,1,0,0,0,... >> 4 = 2,0,0,0,0,... >> 5 = 0,0,1,0,0,... >> ... >> 28 = 2,0,0,1,0,0,0,... >> ... >
>> D = 1,1,1,1,1,... > > > I can see that one may complain that D is clearly infinite and
> therefore should not be in the set, ... Yes. D does not describe a natural number.
> ... but consider the following... > > Let's take the original set and reorder it by exchanging the places > of the i'th prime number with that of the number in the i'th > position. (i.e. First switch the number 2 with the number 1 to move > it to the first position. Then switch 3 with the number -- now 1 -- in > the 2nd position. Then 5 with the 1 which is now in the 3rd position. > Etc...) Because we are just trading the positions of the numbers, all > the same numbers will be in the set afterwards. > > The set is now: >> 2 = 1,0,0,0,0,... >> 3 = 0,1,0,0,0,... >> 5 = 0,0,1,0,0,... >> 7 = 0,0,0,1,0,... >> 11= 0,0,0,0,1,... >> ...
What does "..." mean here? It seems to me you are just enumerating the prime numbers. At which step will you put the numbers 1, 4, 6, etc. If you do have a way to add, in the limit, those numbers in the set, then you are just constructing an order type (ordinal) bigger than omega on the set of positive integers. But such constructive) ordinal are all enumerable.
You could have said directly: let us consider the following order: it is the usual order except that we decide that all even numbers are bigger than the odd numbers, so we have the order: 1, 3, 5, 7, 9, .... 2, 4, 6, 8, 10, .... This is an example of order type OMEGA+OMEGA So what? We can easily enumerate it by zigzagging between the even and odd numbers. > >> D = 0,0,0,0,... > > > I would suggest that the diagonal method does not find a number which > is different from all the members of a set, but rather finds a number > which is infinitely far out in the ordered set.
Like I say. Your construction put a bigger, but still enumerable, order on N. Actually we have already used diagonalization for building constructive ordinal, or order type on the set of computable growing functions. But those produce enumerable sets.
> > If anyone can find where I've gone wrong, please let me know.
Cantor showed that ALL tentative enumeration of some set S fails. This is what makes that set S non enumerable. You are just showing that the diagonal method can also work on some precise enumeration on N. This does not make N non enumerable, it makes only your precise enumeration incomplete (or extendible in the constructive ordinal, but that does not change anything). A set is non enumerable if ALL attempts to enumerate it fail, not if some particular attempt fails. That is why we do a proof by a reductio ad absurdo.
In you next post you say (to Barry): > Let me see if I am clear about Cantor's method. Given a set S, and > some enumeration of that set
Well S is fixed. It is the set we want to show being NOT enumerable. Then you take not some enumeration, but ANY enumeration.
> (i.e., a no one-one onto map from Z^+ to S) we can use the > diagonalization method to find an D which is a valid element of S but > is different from any particular indexed element in the enumeration.
... is different from any particular indexed element, for any arbitrary enumeration. You have to be sure that the diagonal will work whatever enumeration is proposed.
> Cantor's argument then goes on to say (and here is where I disagree > with it) that therefore D is not included in the enumeration and the > enumeration is incomplete. > > I, on the other hand, would posit that the enumeration may include > elements whose index is not well defined.
Hmmmm.... In Cantor's proof of the non enumerability of 2^N (the set of infinite binary sequences), the indexes are all perfectly well defined even if it is in Platonia or in the mind of God, so your remark does not apply.
But now, curiously enough, a remark similar to your's can be done about the proof that the (total) computable functions from N to N are not *computably* enumerable. In that case, if we assume Church thesis, and thus the existence of a universal language L, the set of all (total) computable functions from N to N *is* enumerable, but is not computably enumerable. The lesson is that there is a bijection between the set of indexes of the total computable functions from N to N, and N, but that bijection is not a computable function. See the preceding "key" post, because you are perhaps confusing effective (computable) enumeration and non effective enumeration. I recall that the diagonal argument, when applied on the set of all functions (from N to N) proves that that set is not enumerable, but when the diagonal argument is applied to the (obviously) enumerable set of computable functions (from N to N) it shows that the enumeration (which exists) is not a computable one. Exercise: What is wrong with the following argument. (I recall that by definition a function from N to N is defined on all natural numbers). (false) theorem: the set of computable functions from N to N is not enumerable. (erroneous) proof: let us suppose (by absurdum) that the set of computable functions from N to N is enumerable. Thus there exist an enumeration f_1, f_2, f_3, ... f_i, .... of the computable functions from N to N. But then I can define the following computable function g, from N to N, by: g(n) = f_n(n) + 1 g is computable: to compute it just search in the enumeration the function f_n, which is computable, and then apply it on n, and then add 1. But then g has to be in that enumeration f_i of the computable function. Thus there is a k such that g = f_k. In particular g(k) = f_k(k). But g(k) = f_k(k) + 1, by definition of g. Thus f_k(k) = f_k(k)+1. And the f_i are computable functions, so f_k(k) is a well defined number, which I can subtract on the left and the right side of f_k(k) = f_k(k)+1, giving 0 = 1. Contradiction. So the set of computable functions from N to N is not enumerable. What is wrong? We know that the set of computable function has to be enumerable, because "computable" means we can describe how to compute the function in a finite expression in some language, and the set of all finite expressions has been shown enumerable. So what happens? Bruno http://iridia.ulb.ac.be/~marchal/
|
| Re: Cantor's Diagonal |
Brent |
17/12/07 10:04 |
Bruno Marchal wrote: > Hi Daniel, > > I agree with Barry, but apaprently you have still a problem, so I > comment your posts. > > > Le 16-déc.-07, à 10:49, Daniel Grubbs a écrit : > > Hi Folks, > > I joined this list a while ago but I haven't really kept up. > Anyway, I saw the reference to Cantor's Diagonal and thought perhaps > someone could help me. > > Consider the set of positive integers: {1,2,3,...}, but rather than > write them in this standard notation we'll use what I'll call 'prime > notation'. > > > > > OK. That is often used to compare the purely additive and the purely > multiplicative structures of the natural numbers. Of course the number 0 > is banished in the multiplicative structure (so you have to handle 0 > manualy, but that does not change the enumeration question, ok). > > > > > Here, the number m may be written as > > m = n1,n2,n3,n4,... > >
> 1 = *0*,0,0,0,0,... > 2 = 1,*0*,0,0,0,... > 3 = 0,1,*0*,0,0,... > 4 = 2,0,0,*0*,0,... > 5 = 0,0,1,0,*0*,... > ... > 28 = 2,0,0,1,0,0,0,... > ... > > > D = 1,1,1,1,1,... > > > > I can see that one may complain that D is clearly infinite and > therefore should not be in the set, ... > > > > Yes. D does not describe a natural number. > > > > ... but consider the following... > > Let's take the original set and reorder it by exchanging the places > of the i'th prime number with that of the number in the i'th > position. (i.e. First switch the number 2 with the number 1 to move > it to the first position. Then switch 3 with the number -- now 1 -- > in the 2nd position. Then 5 with the 1 which is now in the 3rd > position. Etc...) Because we are just trading the positions of the > numbers, all the same numbers will be in the set afterwards. > > The set is now: >
> 2 = *1*,0,0,0,0,... > 3 = 0,*1*,0,0,0,... > 5 = 0,0,*1*,0,0,... > 7 = 0,0,0,*1*,0,... > 11= 0,0,0,0,*1*,... If you tried to compute g(k) and g was in the list, i.e. some f_k, then when you searched a for g, when you came to f_k it would start with the prescription "...just search in the enumeration..." and you would be in a infinite loop. Brent Meeker
|
| Re: Brent's answer to Bruno's puzzle |
Barry Brent |
17/12/07 23:31 |
Brent, I don't think this is enough. There may be a different algorithm for f_k that escapes your infinite loop. If this alternative algoritm doesn't exist, I think you need to show that it doesn't exist. I suspect that the error in Bruno's erroneous proof has to do with formal languages. Say a function f is computable with regard to a formal language L when there exists an algorithm written in L to compute f. The f_n are computable with regard to some particular formal language. Functions computable with regard to one language may or may not be computable with regard to another language. (If this is false, Bruno needs to prove it's false.) Thus when Bruno argues that the set of computable functions is enumerable, he must mean "for any fixed language L, the functions computable with regard to L is enumerable." Now the procedure Bruno described for computing g is not written in a formal language--it's written in English. The fact that this English language description is finite doesn't prove that g is computable with regard to L, ie, doesn't prove that g is one of the f_n. I'm an amateur at this--this "solution" is really just a question for Bruno... Barry (Barry Brent, not Brent Meeker!)
|
| Re: Cantor's Diagonal |
Bruno Marchal |
18/12/07 06:49 |
Le 17-déc.-07, à 19:04, meekerdb (Brent Meeker) wrote: This is a bit fuzzy, but then the exercise was a bit fuzzy too, so I can considered this as correct. More below. Barry Brent wrote:
> Brent, I don't think this is enough. There may be a different > algorithm for f_k that escapes your infinite loop. If this > alternative algoritm doesn't exist, I think you need to show that it > doesn't exist. > > I suspect that the error in Bruno's erroneous proof has to do with > formal languages. Say a function f is computable with regard to a > formal language L when there exists an algorithm written in L to > compute f. The f_n are computable with regard to some particular > formal language. Functions computable with regard to one language > may or may not be computable with regard to another language. (If > this is false, Bruno needs to prove it's false.) Thus when Bruno > argues that the set of computable functions is enumerable, he must > mean "for any fixed language L, the functions computable with regard > to L is enumerable." Now the procedure Bruno described for computing > g is not written in a formal language--it's written in English. The > fact that this English language description is finite doesn't prove > that g is computable with regard to L, ie, doesn't prove that g is > one of the f_n. > > I'm an amateur at this--this "solution" is really just a question for > Bruno... > > > Barry (Barry Brent, not Brent Meeker!) Now, if this comment was correct, it would mean that universal languages or universal machines would not exist!
Actually, I can also take this remark as a correct conclusion, but only by considering a restricted notion of universal machines or language. Barry, are you actually denying Church Thesis (which says that an universal language exists, indeed LAMBDA-CALCULUS is one!) ? Barry Brent, Brent Meeker, you are both correct, individually, but you cannot be both correct simultaneously, so what? I let you and others think a bit more, for the fun of it, for the importance of being 100% clear on this before proceeding, and because I have not the time to say more today. Don't worry, even the biggest one like Godel, Kleene, Church ... all have been wrong on this at some moment. It *is* a bit subtle. But all the possibility of my work, or more generally, the very consistency of comp relies on that subtlety. Only after a good understanding on this, will I be able to come back on less technical explanation. If I explain things non technically to soon I will have the feeling to "take you in a boat" (we say in French), meaning roughly to be dishonest. It is worth to take the time, and to play a bit with the idea, right? Bruno PS I know also, for having explained this years ago in the list that some in the list does understand what happen here, but I encourage them to go through the reasoning again, because I am not sure they did understand the *impact* of this .... It is one thing to understand a proposition, and another thing to get the importance, relatively to the TOE search of what is really going on here ... In particular, when the ASSA people invoke Kolmogorov complexity notions, they do rely on Church Thesis .... The "key post" is a key for everybody, I mean for both ASSA and RSSA type of TOE researchers. http://iridia.ulb.ac.be/~marchal/ |
| Re: Cantor's Diagonal |
Barry Brent |
18/12/07 09:52 |
Bruno-- Ahh, my amateur status is nakedly exposed. I'm going to expose my confusion even further now. Never heard of a universal language. I thought I was familiar with Church's thesis, but apparently no. I thought it was the claim that two or three or four concepts (including recursive function and computable function) were extensionally equivalent. I have heard of the lambda calculus, but I don't know what it is, or what its connection is with Church's thesis. I have a rough guess, based on what you're saying. I'm surprised. I imagine that the claim of existence of a universal language must be made in the context of a theory of languages? Never heard of that theory. Well, if I imagine such a theory, it must involve both syntax and semantics, yes? Semantics connects a language to a world, right? (The experts are cringing, I'm sure....) Can one language encompass all possible worlds? Can't we imagine worlds, the structures of which are so dissonant, that their languages could never be consistently subsumed under some single larger ("universal") language? (More cringing, no doubt....) Or, is it that when we restrict the worlds in question to some suitable realm--say, numbers-- all these things work out? (Cringes redoubled!) I can imagine other ways out. Maybe we're concerned with just one world, suitably described. Maybe structural inconsistencies of possible worlds are no more an impediment to being expressible in one language than logical inconsistencies? (But how do we know?) What about cardinality? From your remarks, I imagine that the number of elementary symbols in any language, including the universal language, is supposed to be finite, so that the set of algorithms is countable? If there are lots of worlds and languages, I wonder how people make that work. Is such a language going to be adequate for expressing propositions about all possible worlds? How do we know? My poor excuse is, I've only been on this list a little while, and I made my (sketchy) acquaintance with these ideas a very long time ago. Sorry if I'm way off topic. Tell me to go look it up somewhere, or stop wasting time, if you want to... Barry Brent Dr. Barry Brent barry...@earthlink.net http://home.earthlink.net/~barryb0/ |
| Re: Cantor's Diagonal |
Bruno Marchal |
19/12/07 05:57 |
Hi Barry, Le 18-déc.-07, à 18:52, Barry Brent a écrit :
> > Bruno-- > > Ahh, my amateur status is nakedly exposed. I'm going to expose my > confusion even further now. That is the courageous attitude of the authentic scientists. I like "amateur" because they have less prejudices, they have inner motivations, and rarely follow authoritative arguments.
> > Never heard of a universal language. I thought I was familiar with > Church's thesis, but apparently no.
As I said a while ago, coming back from an international meeting on computability (in Siena, where my Plotinus' paper has been accepted), I got the feeling that few people really grasp Church Thesis, including some "experts". My "Brussels thesis" has been criticized for being too much pedagogical on Church thesis, but each time people try to debunk my work, soon enough I realize they have a problem with Godel or Church, never (yet) with my contribution. The worst is that most people *feel* at ease with CT but apparently are not. I take as an honor to explain this too you, I *do* appreciate your work in Number Theory, as far as I understand it. Possible links could emerge. (You make me discover also the nice paper on "prime percolation" by Vardi: I love percolation. Not just because I am an amateur of good coffe, but because exact percolation problem have led to the Temperley Lieb algebra/category; which makes links between knot theory, combinators/lambda-calculus, quantum computations, and eventually number theory, if not the number 24 itself).
> I thought it was the claim that > two or three or four concepts (including recursive function and > computable function) were extensionally equivalent. I have heard of > the lambda calculus, but I don't know what it is, or what its > connection is with Church's thesis. I have a rough guess, based on > what you're saying. I'm surprised. I imagine that the claim of > existence of a universal language must be made in the context of a > theory of languages? Never heard of that theory. This happens because the expression "theory of languages" is used in the context of "non universal languages", like in the Chomsky hierarchy of languages for example. Universal languages and machines appears in what is called "computability theory" or "recursion theory".
> > Well, if I imagine such a theory, it must involve both syntax and > semantics, yes? Semantics connects a language to a world, right? > (The experts are cringing, I'm sure....) Can one language encompass > all possible worlds? Can't we imagine worlds, the structures of > which are so dissonant, that their languages could never be > consistently subsumed under some single larger ("universal") > language? (More cringing, no doubt....) Or, is it that when we > restrict the worlds in question to some suitable realm--say, numbers-- > all these things work out? (Cringes redoubled!) I can imagine other > ways out. Maybe we're concerned with just one world, suitably > described. Maybe structural inconsistencies of possible worlds are > no more an impediment to being expressible in one language than > logical inconsistencies? (But how do we know?)
We have to distinguish logic and computability. In logic we will have language in which sentences are to be interpreted in some world/model. But in computability we can go very far by just interpreting them in some procedural way. The expressions in computer language are really basic instruction like in the coffee-bar machine. Eventually we can describe them all in term of NAND gates, delay and electrical current. A computing language is then universal if all computable functions (from N to N, or from finite things coded in N to finite things coded in N) can be computed by following a finite set of instructions in the language.
> > What about cardinality? From your remarks, I imagine that the number > of elementary symbols in any language, including the universal > language, is supposed to be finite, Yes.
> so that the set of algorithms is > countable? Yes. And so are the computable functions.
> If there are lots of worlds and languages, I wonder how > people make that work.
Because all the languages or machines which have been invented for computing (computable) functions from N to N, have been shown equivalent, and that the closure of the set of computable functions by those machines, for the (transcendental) diagonalization procedure, give a powerful argument that those language/machine are universal. Careful: they are universal with respect to the class of computable functions. They are not universal with respect to the propositions they can express or prove. A universal language/machine is not a theory. Most universal language don't even have a way to assert propositions, just some sort of commands (cf the coffee-bar instructions as example).
> Is such a language going to be adequate for > expressing propositions about all possible worlds? How do we know? In computing, well ... like in Number theory, there is only one "world", the world of natural numbers. But universal language just describes functions from N to N. Propositions will occur when we will introduce basic beliefs in the machine. Such beliefs will NEVER be universal. Universality is a computability notion. By Godel, no theories can be universal with respect of provability.
> > My poor excuse is, I've only been on this list a little while, and I > made my (sketchy) acquaintance with these ideas a very long time > ago. Sorry if I'm way off topic. I think you are quite in the topic. To explain a bit of my work (and related TOE things) I have to provide some help for people to distinguish clearly many things which are quite different but which are also deeply related. It is the relation in between those notions which are important. They are often confused. Those notions are (forgetting the most important one "truth"):
- Computability (the only one which has a possible notion of universality) - Provability (which *can* be universal with respect to computability, but which is never universal with respect to provability of propositions). This will be illustrated with the notion of "lobian machine". - Knowability (which can be proved equivalent extensionally with provability, but will appears to have quite different intensional logics) - Observability (the same remark applies). - Sensibility Etc. More later ... I will come back on the "key post" asap. > Tell me to go look it up > somewhere, or stop wasting time, if you want to... Actually I'm afraid your are just motivating me for trying to be clearer and simpler. You are the one helping here, but feel free to organize your computability-time as you have to. Best,
Bruno http://iridia.ulb.ac.be/~marchal/ |
| Re: Cantor's Diagonal |
Barry Brent |
19/12/07 12:09 |
|
| Re: Cantor's Diagonal |
Bruno Marchal |
20/12/07 07:28 |
Le 19-déc.-07, à 21:09, Barry Brent a écrit : > > Excellent, Bruno, Thanks! Well thanks. I will send a next diagonalization post and some references next week, Best, Bruno http://iridia.ulb.ac.be/~marchal/ |
| Re: Cantor's Diagonal |
Günther Greindl |
21/12/07 04:08 |
Hi Russell, Russell Standish wrote: > In your first case, the number (1,1,1,1...) is not a natural number, > since it is infinite. In the second case, (0,0,0,...) is a natural > number, but is also on the list (at infinity). Why is (1,1,1,...) not in the list but (0,0,0,...) in the list at infinity? This seems very arbitrary to me. I am becoming more and more an ultra-finitist. Arguments with infinity seem to be very based on the assumptions you make (about platonia or whatever) Regards, Günther -- Günther Greindl Department of Philosophy of Science University of Vienna guenther...@univie.ac.at http://www.univie.ac.at/Wissenschaftstheorie/
Blog: http://dao.complexitystudies.org/ Site: http://www.complexitystudies.org
|
| Re: Cantor's Diagonal |
Quentin Anciaux |
21/12/07 04:22 |
Hi, Le Friday 21 December 2007 13:08:38 Günther Greindl, vous avez écrit : > Hi Russell, > > Russell Standish wrote: > > In your first case, the number (1,1,1,1...) is not a natural number, > > since it is infinite. In the second case, (0,0,0,...) is a natural > > number, but is also on the list (at infinity). > > Why is (1,1,1,...) not in the list but (0,0,0,...) in the list at > infinity? This seems very arbitrary to me. Because zero even repeated an infinity of time is zero and is a natural number. (1,1,1,...) can't be a natural number because it is not finite and a natural number is finite. If it was a natural number, then N would not have a total ordering. > I am becoming more and more an ultra-finitist. Arguments with infinity > seem to be very based on the assumptions you make (about platonia or > whatever) Finite and infinite concepts are dual concepts you can't leave one without leaving the other. > Regards, > Günther Regards, Quentin Anciaux
-- All those moments will be lost in time, like tears in rain.
|
| Re: Cantor's Diagonal |
Günther Greindl |
21/12/07 04:30 |
Hi, > Because zero even repeated an infinity of time is zero and is a natural > number. (1,1,1,...) can't be a natural number because it is not finite and a > natural number is finite. If it was a natural number, then N would not have a > total ordering. Ok agreed: I was caught up in viewing it simply as an "indexing" scheme, but viewed constructively I of course agree. My error. >> I am becoming more and more an ultra-finitist. Arguments with infinity >> seem to be very based on the assumptions you make (about platonia or >> whatever) > > Finite and infinite concepts are dual concepts you can't leave one without > leaving the other. Could you elaborate some more on this? Regards, Günther -- Günther Greindl Department of Philosophy of Science University of Vienna guenther...@univie.ac.at http://www.univie.ac.at/Wissenschaftstheorie/
Blog: http://dao.complexitystudies.org/ Site: http://www.complexitystudies.org
|
| Re: Cantor's Diagonal |
Russell Standish |
23/12/07 15:08 |
On Fri, Dec 21, 2007 at 01:08:38PM +0100, Günther Greindl wrote: > > Hi Russell, > > Russell Standish wrote: > > > In your first case, the number (1,1,1,1...) is not a natural number, > > since it is infinite. In the second case, (0,0,0,...) is a natural > > number, but is also on the list (at infinity). > > Why is (1,1,1,...) not in the list but (0,0,0,...) in the list at > infinity? This seems very arbitrary to me. Quentin replied correctly on the first point. On the second point, the list is assumed to contain all natural numbers. If that is true, the 0, which maps to the sequence (0,0,0...) must lie at infinity on the list. If you don't accept this, then the enumeration given is not an enumeration of the naturals, since 0 is then not on the list. Either way falsifies the argument. > > I am becoming more and more an ultra-finitist. Arguments with infinity > seem to be very based on the assumptions you make (about platonia or > whatever) > > Regards, > Günther > > > -- > Günther Greindl > Department of Philosophy of Science > University of Vienna > guenther...@univie.ac.at > http://www.univie.ac.at/Wissenschaftstheorie/ > > Blog: http://dao.complexitystudies.org/ > Site: http://www.complexitystudies.org > >
-- ---------------------------------------------------------------------------- A/Prof Russell Standish Phone 0425 253119 (mobile) Mathematics UNSW SYDNEY 2052 hpc...@hpcoders.com.au Australia http://www.hpcoders.com.au ----------------------------------------------------------------------------
|
| DIAGONAL (answer to the problem) |
Bruno Marchal |
31/12/07 05:38 |
Hi, I said: Answer :
> > (false) theorem: the set of computable > > functions from N to N is not enumerable.
Yes that is wrong. The set of computable functions from N to N is enumerable.
> > (erroneous) proof: OK, let us see where we are wrong.
> > let us suppose (by absurdum) that the set of > > computable functions from N to N is enumerable. OK. (although we already know that the set of computable functions from N to N is enumerable, here it is the intended absurd hypothesis).
> > Thus there exist an > > enumeration f_1, f_2, f_3, ... f_i, .... of > > the computable functions from N to N.
Yes, sure. That is what is meant by enumerable.
> > But then I can define the following > > computable function g, from N to N, by: > > > g(n) = f_n(n) + 1 > > > g is computable: to compute it just search > >in the enumeration the function f_n, which is > >computable, and then apply it on n, and then add 1.
Here is the error. From the existence of an enumeration f_0, f_1, .... of the computable functions from N to N, the existence of a mechanical, algorithmic, computable enumeration of those f_i does not follow. And this is needed for making g computable from N to N. Although each f_i is a computable function, it is the correspondence i ----- f_i which is not computable, making g NOT computable either. There is indeed a bijection between N and the set of computable functions from N to N, but that bijection is not computable. It has to be! All right? GENERAL COMMENT: What are the lessons of those diagonalizations? First I will assume a weak form of Church thesis WCT: WCT: There exist a universal language (machine) By definition, in such language, all computable functions have an expression understandable by the corresponding universal machine. We can then enumerate the expressions in the universal language: F_1 F_2 F_3, ..., but, by the theorem above, the list of those expressions has to enumerate a larger set than the set of computable functions from N to N. We get actually an enumeration of all computable functions from some subset of N to N. The so-called computable partial functions from N to N. A function from N to N is a particular case of a function from some subset of N to N, given that N is a subset of N (!), so the class of the functions from N to N is included in the class of the partial functions from N to N. If the domain of definition of f is different from N I will say that f is strictly partial. To emphasize that f is not strictly partial, I will say that f is total: it just means f is defined on all n. We learn more: there is no expression in the universal language capable of deciding if n is the code of a f_n being total or strictly partial. For if that would exist then we would have a procedure for extracting from the F_i the f_i, making the function g above computable, and we know it can't. We also get a very strong incompleness theorem: for all effective theory there are statement, like "n codes a total function" which are undecidable in the theory. Indeed, if there was an effective theory capable of deciding all statements of the type :"n codes a total function" then such a theory would provide a way to build an expression in the language making possible to enumerate the total computable functions, which can't be. So, at a fundamental level, it shows that any machine, capable of changing itself, has to choose between universality and security. If you want all the total computable functions, not only your machine can not stop on some inputs, but it does it sometimes in a non effectively provable of verifiable way. Giving the enumerabilty of the (total or not) computable functions, which follows directly from CT, Cantor diagonalization already shows the existence of non computable functions from N to N. Indeed most of them are not computable. The set of computable functions is enumerable but the set of all functions is not enumerable, making the set of uncomputable functions from N to N not enumerable. In that sense there are far more non computable functions from N to N than computable one. But the effective diagonalization, made on the partial computable functions shows, with CT, that the function TOT, from N to N, such that TOT(n) = 1 if F_n is total, and 0 if not, is not computable. That, is, we get the uncomputability of precise functions and predicates bearing in computer science (and by Godel's arithmetization, already in number theory). ... IS WCT true? Well CT -> WCT, and there are many evidences for CT. (ASAP ...) Hope this provide some help.
Happy new year, Bruno http://iridia.ulb.ac.be/~marchal/ |
| Re: coffee-bar machine excerices |
Mirek Dobsicek |
07/01/08 07:51 |
BEGIN 1: Z(1) 2: Z(2) 3: J(1,2,3) # true condition, loop END > 2) find a job which computes addition (which is of course a function > from NXN to N)
BEGIN 1: Z(3) # clean temp tables 2: Z(4) 3: J(2,3,8) # are we done? get out of the loop 4: S(1) 5: S(3) 6: S(4) 7: J(3,4,3) # always true condition, continue with adding 8: END > 3) using the preceding job, find a job which computes multiplication. BEGIN 1: Z(5) # clean temp tables 2: Z(6) 3: T(1,5) # copy 1 -> 5 4: J(5,6,14) # are we done? get out of the multipl. loop 5: S(6) 6: Z(3) # adding loop start 7: Z(4) 8: J(2,3,13) 9: S(7) 10: S(3) 11: S(4) 12: J(3,4,8) # adding loop end 13: J(3,4,4) # always true condition, continue with multipl. loop 14: Z(1) 15: T(7,1) # copy 7 -> 1, table one contains the final result END > 4) is the following proposition plausible: a function from N to N is > intuitively computable if and only if it can be computed by some coffee > bar job. yes 1) instructions of the coffee-bar language coincide with important instructions of nowadays central processing units in computers 2) coffee-bar instructions are sufficient for constructing logical AND, OR, NOT operations 3) I should be able to find a better reason, damn... > 5) describe informally the coffee-bar language, and, choosing an order > on its alphabet, write the first 7 jobs in the lexicographical order. > The alphabet contains all symbols needed in the jobs, including commas, > parentheses, etc. + some grammatical rules making clear that Z(23) is a > good instruction, but 23(Z) is not, ... Program ::= BEGIN commands END commands ::= line-no instruction comment next line-no ::= num: instruction ::= Z(num) | S(num) | T(num,num) | J(num,num,num) comment ::= # text `new-line` | `new-line` next ::= commands | END num ::= [0,1,2,3,...] text ::= [ascii text without the `new-line` character] First 7 jobs
1) BEGIN 0: J(0,0,0) END 2) BEGIN 0: J(0,0,1) END 3) BEGIN 0: J(0,1,0) END 4) BEGIN 0: J(0,1,1) END 5) BEGIN 0: J(1,0,0) END 6) BEGIN 0: J(1,0,1) END 7) BEGIN 0: J(1,1,0) END Assuming empty tables, programs 1,3,5,7 never reach END. Sincerely, Mirek
|
| Re: coffee-bar machine excerices |
Bruno Marchal |
09/01/08 07:53 |
Le 07-janv.-08, à 16:51, Mirek Dobsicek a écrit : > > Bruno wrote: Excellent. Supposing that ":" has been included in the alphabet. Good idea: it is more readable. You have also add commentaries, beginning by "#". This is of course not part of the formal job/program. Also, note that the original Cutland-Shepherdson Sturgis programs are not self-delimiting (no BEGIN nor END, but I use self-delimiting program most often, so ok).
Here is simpler solution: BEGIN 1: J(1,1,1) # this will work independently of the number of coffee cups already on the table! END > > >> 2) find a job which computes addition (which is of course a function >> from NXN to N) > > BEGIN > 1: Z(3) # clean temp tables > 2: Z(4) > 3: J(2,3,8) # are we done? get out of the loop > 4: S(1) > 5: S(3) > 6: S(4) > 7: J(3,4,3) # always true condition, continue with adding > 8: > END Very good. You could have make a simpler one too, just adding one cup of coffee on table 1 and table 3 and stopping when table 2 and table 3 have the same number, but this is detail. Note that the 8th instruction is not needed and should be suppressed. Programs ends when they does not find the next instruction. "8:" is a bit ambiguous.
> >> 3) using the preceding job, find a job which computes multiplication. > > BEGIN > 1: Z(5) # clean temp tables > 2: Z(6) > 3: T(1,5) # copy 1 -> 5 > 4: J(5,6,14) # are we done? get out of the multipl. loop > 5: S(6) > 6: Z(3) # adding loop start > 7: Z(4) > 8: J(2,3,13) > 9: S(7) > 10: S(3) > 11: S(4) > 12: J(3,4,8) # adding loop end > 13: J(3,4,4) # always true condition, continue with multipl. loop > 14: Z(1) > 15: T(7,1) # copy 7 -> 1, table one contains the final result > END
Perfect. The others should at least see the addition program above used as a "subroutine", here. > >> 4) is the following proposition plausible: a function from N to N is >> intuitively computable if and only if it can be computed by some >> coffee >> bar job. > > yes > 1) instructions of the coffee-bar language coincide with important > instructions of nowadays central processing units in computers
OK. But that is not an argument. The subset with the instructions Z, S, T also, but is not universal. But I see your point: the main fundamental instructions of central processing units belongs to the coffee-bar computer.
> > 2) coffee-bar instructions are sufficient for constructing logical AND, > OR, NOT operations AND and NOT are already universal indeed, if you add a delay instruction (enginners does not say this explictly because the "natural-concrete" AND introduces automatically a delay. And also, you need the "duplicate" instruction, (a bifurcating electrical wire) which engineers mention only since ... quantum computing exists. This is because a bifurcating wire seems a so obvious "instructions" that we forget to mention it, but in quantum computing you just cannot duplicate arbitrary information .... But OK, good idea. > > 3) I should be able to find a better reason, damn... It is not obvious. To be sure it is universal, the better way would be to show you can implement an already known universal system, or to verify explicitly the closure for the diagonal + the emulability of elementary functions, etc. > >> 5) describe informally the coffee-bar language, and, choosing an order >> on its alphabet, write the first 7 jobs in the lexicographical order. >> The alphabet contains all symbols needed in the jobs, including >> commas, >> parentheses, etc. + some grammatical rules making clear that Z(23) is >> a >> good instruction, but 23(Z) is not, ... > > Program ::= BEGIN commands END > commands ::= line-no instruction comment next > line-no ::= num: > instruction ::= Z(num) | S(num) | T(num,num) | J(num,num,num) > comment ::= # text `new-line` | `new-line` > next ::= commands | END > num ::= [0,1,2,3,...] > text ::= [ascii text without the `new-line` character] > > > First 7 jobs > > 1) > BEGIN > 0: J(0,0,0) > END Hmmm.... First I strongly suggest no putting the commentaries in the program (# ...). You just make your life harder.
Second, if you accept, like above a empty instruction (like the 8: in your addition program above, you have to agree that BEGIN 0: END is shorter, no? But I disallowed it. But then why is not the following program BEGIN 0: S(0) END shorter? And third, do you know a coffee-bar with a table numbered by 0? The same for the numbering of the instructions in the job. You have yourself begun the numbering of the instructions with 1? (a total detail, sure). OK. Very good. I hope others understand what you have done. Perhaps someone can correct the last exercise, or perhaps you have a reason to consider J(1,1,1) little than Z(1), which I have not seen?
I give two new exercises: 1) the alphabet of the coffee-bar language (accepting your ":" improvement) is {:, (, ), J, S, T, Z, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9} (we write the number in decimal). Now a program, like BEGIN 1: S(1) END is the same as BEGIN1:S(1)END (less readable, but it is the same program: we are not in PYTHON where the carriage return is in the alphabet!). But the expression BEGIN1:S(1)END denotes a number in base 17. All right? The question is: does a program U exists which is such that if you put n, written in base 17, output nothing on table 1 in case the number is not a program, and output the result of the running of that program in the other case. Would you say that such a U program is universal? 2) Here is another universal system.
Alphabet = { ), (, S, K}. The alphabet has only four elements. Programs/jobs in this system are easy to define inductively: K is a program S is a program if X and Y are programs, then (XY) is a program. Nothing else is a program. So for example the first programs are K, S, (KK) (KS) (SK) (SS) (K(KK)) (K(KS)) (K(SK)) (K(SS)) ... the operational interpretative rule (the dynamic if you want) is given by the following reduction rule replace any occurence of ((Kx)y)) by x replace any occurence of (((Sx)y)z) by ((xz)(yz)); x y z being arbitrary programs. It is easier to read those expressions if you eliminate all left parenthesis: this gives for the dynamical rule: Kxy = x Sxyz = xz(yz) the question are the same than the one for the coffee-bar. In particular could you write a program which crash the system? Hint: search for a program I such that Ix = x. Search a program M such that Mx = xx. Raahhhh ... Sorry but I am interrupted. I come back on this later, Best, Bruno http://iridia.ulb.ac.be/~marchal/ |
| Re: coffee-bar machine exercises |
Mirek Dobsicek |
30/01/08 04:42 |
Well :-) my lexicographical ordering abilities have shown a weak point. > I give two new exercises: > > 1) the alphabet of the coffee-bar language (accepting your ":" > improvement) is > > {:, (, ), J, S, T, Z, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9} (we write the > number in decimal). > > Now a program, like > > BEGIN > 1: S(1) > END > > is the same as BEGIN1:S(1)END (less readable, but it is the same > program: we are not in PYTHON where the carriage return is in the > alphabet!). > > But the expression BEGIN1:S(1)END denotes a number in base 17. All > right? Yes, it is clear. > The question is: does a program U exists which is such that if you put > n, written in base 17, output nothing on table 1 in case the number is > not a program, and output the result of the running of that program in > the other case. Would you say that such a U program is universal? Yes. We call such a program an interpreter. And the existence of a program called 'interpreter' is a consequence of the fact that a machine capable of executing a universal language L, must be descriable itself by L. What happens when we feed U with its own code? Doing something, doing something and .. hanging ... waiting for an input. Best, Mirek
|
| Re: Bijections (was OM = SIGMA1) |
Mirek Dobsicek |
30/01/08 04:43 |
Bruno Marchal wrote: > > Le 20-nov.-07, à 12:14, Torgny Tholerus a écrit : > >> Bruno Marchal skrev: >>> To sum up; finite ordinal and finite cardinal coincide. Concerning >>> infinite "number" there are much ordinals than cardinals. In between >>> two different infinite cardinal, there will be an infinity of ordinal. >>> We have already seen that omega, omega+1, ... omega+omega, >>> omega+omega+1, ....3.omega, ... 4.omega .... ....omega.omega ..... >>> omega.omega.omega, .....omega^omega ..... are all different ordinals, >>> but all have the same cardinality. >>> >> Was it not an error there? 2^omega is just the number of all subsets >> of >> omega, and the number of all subsets always have bigger cardinality >> than >> the set. > > > Yes, that is true. > > > >> So omega^omega can not have the same cardinality as omega. > > > But addition, multiplication, and thus exponentiation are not the same > operation for ordinals and cardinals. I should have written > omega"^"omega, or something like that. That is why I have written > 3.omega instead of 3*omega. Uu, reading about cardinals and ordinals on Wikipeadia did not helped me at this point. Could you please elaborate more on this? Of course, only relatively to its importance towards CT ... Cheers, Mirek |
| Re: Bijections (was OM = SIGMA1) |
Bruno Marchal |
04/02/08 06:21 |
Le 30-janv.-08, à 13:43, Mirek Dobsicek wrote (in different posts): > 2\ Bruno, you recently wrote that you do not agree with Wolfram's > Principle of Computational Equivalence. As I understand to that > principle, Wolfram says that universe is a big cellular automata. What > is the evidence that it is unlikely this way?
Wolfram is very vague about his Principle of Comp Equiv (PCE): http://www.wolframscience.com/nksonline/page-6-text The problem is that Wolfram seems to take a "natural world" for granted, and does not seem to be aware that each of us (us = Lobian machine or even just Lobian entity) cannot discern about 2^ALEPH_0 locally equivalent computations. So Wolfram is either not aware of the consequence of the computationalist hypothesis, or not aware about what we can expect nature to be after knowing that Bell's inequality are violated, quantum measurement "problem" and its MW "solution", etc. Wolfram's idea is inconsistent for the same reason that , "as a TOE" Schmidhuber's constructive approach is inadequate or at least incomplete. Such theories entails COMP. But COMP entails the physical world cannot be entirely a constructive structure. Physics or physicalness emerges from an infinite sum (the nature of which is still under scrutiny) of computational histories, observed in relative perspectives (points of view). There is no reason to believe this leads to a constructive universe. On the contrary, its "geographical and local aspects" could have verifiable non computational feature (like when we repeat spin measurements for example). The problem is that, like many, not only Wolfram and Schmidhuber, seem to take a physical universe for granted, but they take also a sort of identity thesis (brain/mind) for granted. The Universal Dovetailer sequence of thought experiences is supposed to explain in all details why such an identity thesis just can't go through ... Now, Mirek, I don't know if you really want to dig on the UDA and the "philosophy-of-mind/theology" issue, because you can enjoy the math per se. Many people dislike or get stuck in front of the idea that comp makes us duplicable, or by the same token that QM makes us 100% duplicable too, although not 100% clonable. But the UDA does explain why we have to take into account the different points of view (first singular, first plural, third, zeroth, ... ). The UDA, in english, can be found here: The Origin of Physical Laws and Sensations, (Invited Talk SANE 2004). Click on that title, or copy the following in your browser: http://iridia.ulb.ac.be/~marchal/publications/ SANE2004MARCHALAbstract.html (if you study it I would suggest you print the slider too, so that you could perhaps tell me which step you would find hard to go through ....). > Uu, reading about cardinals and ordinals on Wikipeadia did not helped > me > at this point. > > Could you please elaborate more on this? Of course, only relatively to > its importance towards CT ...
I will. Slowly because I will be more and more busy, and also I should write papers, I should extend a bit the Plotinus' one, correct the typo error, submit, etc. I still don't know if it is the physicists, the logicians, or the theologians who will grasp the UDA/AUDA highly interdisciplinary or trans-disciplinary reasoning ... Got not enough feed-back ... I will also solve the combinator "crashing" problem: it does illustrate another form of use of the diagonalization idea ... (this is a hint) ...
Best, Bruno http://iridia.ulb.ac.be/~marchal/
|
| Re: Key Post 1, toward Church Thesis and Lobian machine |
Bruno Marchal |
06/02/08 07:40 |
Le 05-déc.-07, à 23:08, Mirek Dobsicek a écrit : "But thanks to that crashing, *Church thesis remains consistent*. I would just say "An existence of a universal language is not ruled out".
I am ok with you. Consistent (in math) means basically "not rule out". "Formally consistent" means "not formally ruled out", or "not refutable". That is: "Consistent(p") is the same as "~ Provable(~ p)" " ~" = negation like "Provable(p)" is equivalent with "~ Consistent( ~ p)" Some thoughts: Thanks to Godel "completeness" theorem for the first order theory (1930) you can also read consistent(p) by there is a world satisfying p (a world "where" p is true). This relates a syntactical notion (the non existence of a chain of formula derived from the axioms by the use of the inference rules and ending with f) with a semantical: the existence of a mathematical structure satisfying the formula. At least in the frame of many formal classical theories, it is related to the recurrent modal duality: Permitted p <====> ~ Obligatory ~p Obligatory p <====> ~ Permitted ~p
Somewhere p <====> ~ Everywhere ~p Everywhere p <====> ~ Somewhere ~p Sometimes p <====> ~ Always ~p Always p <====> ~ Sometimes ~p Like the usual first order quantifiers: (Ax = for all x; Ex = it exists a x) Ex F(x) <====> ~ Ax ~ F(x) Ax F(x) <====> ~ Ex ~F(x) (all cats are ferocious <====> it does not exist a non ferocious cat) And with formal provability we have also: Consistent p <====> ~ provable ~p Provable p <====> ~ consistent ~p But yes, it is by allowing the machine to crash, and actually by allowing it to crash in a *necessarily* not always predictible way, which makes it possible to be universal.
In a nutshell: Universality ==> insecurity ====> kicking back reality and then (knowledge of your universality) ==> (knowledge of your relative insecurity) ====> (knowledge of a kicking back reality) ===> anticipating an independent "reality" (knowledge of your universality) = lobianity (this I intend to explain later) Mirek asked also in trhe same post:
<<And my last question, consider the profound function f such that f(n) = 1 if there is a sequence of n consecutive fives in the decimal expansion of PI, and f(n) = 0 otherwise Is this an example of a partial computable function?>>
Yes. <<Or is this function as such already considered as un-computable function?>> It could be uncomputable on some value, that is, everywhere the function has value 1, you can in principle compute it (just search the sequence: if it exists you will find it because PI is constructive). If the value is zero, it could be that you will be able to know it, but it could be that you will never know it ...
* * * Something else: Mirek, Brent, Barry, Tom (and all those inclined to do a bit of math): don't read what is following unless you don't want to find the crashing combinators by yourself. I give the solution for the crashing combinators: it is enough to ... mock a mockingbird. Raymond Smullyan calls "mocking bird" a combinator M such that Mx = xx. It is a sort of diagonalisor or duplicator. Now if you apply M on itself, M, that is if you evaluate MM, this matches the left of equation Mx = xx, so MM gives MM gives MM gives MM gives MM ... (crashing!). But does M exists? If you recall well, we know only the existence of K and S, and their descendants: like KK, KS, S(KS), SK(KS)(S(KK)), ... (Recall we don't write any left parenthesis, but something like SK(KS)(S(KK)) really abbreviate the result of applying (SK) to (KS) i.e. ((SK)(KS)) on (S(KK)), i.e. (((SK)(KS))(S(KK))). each combinator can be thought as a function of one variable (itself varying on the combinators). We search a combinator playing the role of M (defined by its behavior Mx = xx). We have only K, S, and their combinations. And we have the two axioms giving the behavior of K and S. Kxy = x K axiom and Sxyz = xz(yz) S axiom Explanation. You can see K as a projector sending (xy) on x, for any y. (imo it is the *subjective* entity per excellence, in particular K discards or eliminate informations like projection does. Church will not allow K or any eliminators in its main systems). Functionally K is Lx Ly . x The variable y is abstracted in some irrelevant way. We want Mx = xx. But xx does not match either x or xz(yz), so that we could use the axioms above directly. But imagine we dispose of the subroutine combinators I such that Ix = x. The identity combinators. Then Mx = xx = Ix(Ix), and this does match xz(yz), so that Ix(Ix) is really SIIx (in Sxyz = xz(yz), so SIIx = Ix(Ix) = xx. So SII can play the role of M, it behaves like M. We could define M by SII. Let us verify MM = SII(SII) does crash the system: SII(SII) = I(SII)(I(SII)) = SII(SII) = I(SII)(I(SII)) = SII(SII) = I(SII)(I(SII)) = SII(SII) = I(SII)(I(SII)) = SII(SII) = ... (crashing). Now we have to still find an identity combinator I such that Ix = x. Now x does match the right of the first axiom Kxy = x. Except that K on x wait for a second argument. So let us give to it a second argument such that we get something matching the second (S) axiom: x = Kx(Kx) = SKKx So SKK does the job. So we can take I = SKK. So M = SII = S(SKK)(SKK) and a crashing expression, sometimes called INFINITY is given by MM = SII(SII) = S(SKK)(SKK)(S(SKK)(SKK)) So, a solution was S(SKK)(SKK)(S(SKK)(SKK)) Remark: Note that an existential quantification "ExP(x)" is a sort of projection too. Eventually, the lobian machine observation-act-decision is just that: projection by elimination of worlds (elimination of accessibility of possibilities, a bit like when you get married, of get a job, etc ....). Bruno http://iridia.ulb.ac.be/~marchal/
|
| Re: Key Post 1, toward Church Thesis and Lobian machine |
JohnM |
06/02/08 14:56 |
Bruno, here is my "out of order and off topic" remark. ---------------------------- We are here in theoretical theorizing by theory-laden theoretic ways. It is ALL the product of a mental exercise. Even a Loebian kick in the ass can be a theoretical halucination. You wrote: "... - ... But does 'M" exist? ,,, - ..." (Never mind in what context. ) "exist" is a hard word. Contemplating in a generalized way, I would say: "Everything (not in Hal's sense) exists what we THINK of, if not otherwise: in our ideas. Does 'K' or 'S' have a better than mental existential veracity? We can think of a symbol that it does or does not exist, it does not change that it DOES indeed exist in our mental domain. Do you have a better 'domain' (e.g. a physical existence)? I doubt. In our 1st person world it would not make sense. ----------------------- Excuse my rambling and please, consider it 'entertainment' rather than discussion-post. John M |
| Re: Key Post 1, toward Church Thesis and Lobian machine |
Bruno Marchal |
07/02/08 07:41 |
John, Le 06-févr.-08, à 23:56, John Mikes a écrit :
> > Bruno, here is my "out of order and off topic" remark. > ---------------------------- > We are here in theoretical theorizing by theory-laden theoretic ways. > It is ALL the product of a mental exercise. Even a Loebian kick in > the ass can be a theoretical halucination.
I could agree. But then *all* theories are hallucinations, all right? Even the baby's theories according to which they have a mother is a theoretical hallucination, most probably emerging from a conversation between billions of connected speculative amoeba/neurons betting on some personal reality ...
> You wrote: > "... - ... > But does 'M" exist? ,,, - ..." > (Never mind in what context. ) > > "exist" is a hard word.
I am not so sure. I mean that in some context the question is clearcut and meaningful (independently of the complexity of solving it). In the current context K and S exists, by definition, and all their descendants (their combinations) exist, by definition too. Now they have all a rather well-defined behavior due to the behavior of K and S, and the question of the existence of M (defined by its duplicator behavior) is becoming a pure engineering problem. Ask me examples if this is not clear.
> Contemplating in a generalized way, I would say: > "Everything (not in Hal's sense) exists what we THINK of, if not > otherwise: in our ideas. Yes sure. Actually K is so perverse (in the eyes of some logicians, like Church) that some wants to say that K does not make sense, and Curry (one of the (re)discover of K) defended the existence of K as an idea of thought. Yes sure: eliminating something is a widespread idea.
> Does 'K' or 'S' have a better than mental existential veracity?
I would suggest you to take a look at my paper "Theoretical computer Science and the natural laws". In that paper I sum up (a bit roughly) the Physics of Newton by "K does not exists (in nature)", and I sum up the Physics of Einstein-Podolski-Rosen-Everett-Deutsch-Zurek-Wooters, by "S does not exist". Indeed K eliminate information (like a "classical black hole") and S duplicates arbitrary informations (a problem in QM). So yes K and S are on the mind side, not on the matter side. But this is not needed in the present context, where I introduced S and K just as an example of programming language (typically already on the mind side).
> We can > think of a symbol that it does or does not exist, it does not change > that it DOES indeed exist in our mental domain.
I don't understand that sentence. (don't confuse the symbol K with the primitive instruction K defined by Kxy = x, it is the left- projection or the right elimination: it send (x, y) on x (eliminating y).
> Do you have a better 'domain' (e.g. a physical existence)? I doubt. > In our 1st person world it would not make sense. Physical existence is, by UDA, at best a first person (plural) construct. I recall you that in "my theory" (my favorite hallucination which I try to share with you) numbers and combinators and alike exists before anything material. Matter emerges as a relative border of the machines/numbers ignorance. I do have a better 'domain': numbers (integers). All the rest are number's hallucinations or first person perspectives. But some hallucination can last lawfully, and the question is why. With comp, the question can be made 100% math, and that makes comp testable. And if you don't like numbers, you could take directly combinators instead (their are just less known for contingent reason like we have digits).
> ----------------------- > > Excuse my rambling and please, consider it 'entertainment' rather than > discussion-post. Your rambling could help me to make things clearer perhaps, but ok, the deepest purpose here is fun and entertainment. Thanks for your attention. Now I will hallucinate a bit on a cup of coffee ...,
Best, Bruno -------------
http://iridia.ulb.ac.be/~marchal/
|
| Re: Key Post 1, toward Church Thesis and Lobian machine |
JohnM |
08/02/08 13:54 |
Bruno, thanx. You play loose with 'context': not observed with the baby's diapers, but observed with K and S - (what I didnot specify at all, in the contrary: I spoke about (any) symbol in the sentence what fou failed to misunderstand rightly. ) You seem to comfortably refer to 'matter' (vs numbers) while we are not on 'physical' basis. (I still cannot fathom how the originating 'numbers' (integers) have the consciousness and will to 'generate' the world. I did not learn that, all I learned was: they are tools "I" can manipulate. (Of course I am no mathematician and 'learned' mostly applied math). One cow has 4 legs, eo ipso 4 cows have 16. It does not solve Hal's Nothing" problem. Your 'numbers' religion still requires a 'numbers God' - or did they bootstrap themselves? This view does not represent a different one to any other what people believe in. I confess freely in my narrative that "...and further BACK we know nothing, I 'imagine' a behavior and take it from there to arrive at the situation we think we see today - using (MY) common sense. I have to see something 'better' than what I use to accept it - instead. I am glad you replied BEFORE your coffe, So I could (more or less) follow it. John M (prior art erased)
|
| Re: Key Post 1, toward Church Thesis and Lobian machine |
Mirek Dobsicek |
11/02/08 08:58 |
> "But thanks to that crashing, *Church thesis remains consistent*. I > would just say "An existence of a universal language is not ruled out". > > > > I am ok with you. Consistent (in math) means basically "not rule out". > "Formally consistent" means "not formally ruled out", or "not refutable". > > That is: > > "Consistent(p") is the same as "~ Provable(~ p)" " ~" = negation > > like "Provable(p)" is equivalent with "~ Consistent( ~ p)" All right... Now, let me just rephrase few points of the key post one more time. I will try to be picky with wording. Points which are not mentioned - I fill comfortable with.
1\ There is no language/algorithm/procedure/machine which can describe/execute all total computable functions. 2\ There exists non-empty set of all machine computable functions (inevitably includes both total and strict partial functions). Let us call this set MC (machine computable). 3\ Church himself *defined* the set of so-far-known-computable-functions as those computable by lambda calculus. 4\ What we use to call a *Church thesis* today says, that MC is in fact equal to the set of functions computable by lambda calculus. 5\ Church thesis is refutable. > * * * > > Something else: >
> to us verify MM = SII(SII) does crash the system: > > SII(SII) = I(SII)(I(SII)) = SII(SII) = I(SII)(I(SII)) = SII(SII) = > I(SII)(I(SII)) = SII(SII) = I(SII)(I(SII)) = SII(SII) = ... (crashing). >
Working with SK combinators, I had a bit problems with omitted parenthesis. Also it was not clear to me what is the meaning of eg. (SKK) since S is expected to take three arguments. What helped me was the unlambda language (http://www.madore.org/~david/programs/unlambda/) So here is my crashing sequence (yep, yours is shorter) (I don't expand the I term to keep it short) SII(SI(S(KI)I)) a reference implementation in unlambda: ```sii``si``s`kii the ` stands for 'apply' operation, aka left parenthesis. with a small modification ```sii``si``s`k.ui we can achieve the computer to print uuuuuuuuuuuuu.... in an endless loop. .u is a special function with a side effect of printing the character u. Best, Mirek |
| UDA paper |
Mirek Dobsicek |
12/02/08 14:20 |
Hi Bruno, > The UDA, in english, can be found here:
> */The Origin of Physical Laws and Sensations/*, (Invited Talk SANE 2004). > Click on that title, or copy the following in your browser: > http://iridia.ulb.ac.be/~marchal/publications/SANE2004MARCHALAbstract.html > (if you study it I would suggest you print the slider too, so that you > could perhaps tell me which step you would find hard to go through ....).
I have started reading this paper. Just a quick question. At the first step of UDA it seems you restrict yourself to classical bits. That is fine. I can imagine that somebody deliberately read and cut my running computer so that the computer goes on with its job after being 'reincarnated' in Helsinky. Even the substitution level is more or less clear. Noise on transistors is definitely not important. However, at the third step you mention quantum mechanics. It is not clear to me how would you classically teleport my quantum computer. What are the read & cut operations?
Yes, there exists a classical Turing machine which can simulate my quantum computer, but I am not giving the running simulator to you. I don't have it. Please, make a short clarification about your framework. I might be just misinterpreting you. What is the page reference to Gruska's book? Sincerely, Mirek
|
| Re: Key Post 1, toward Church Thesis and Lobian machine |
Bruno Marchal |
13/02/08 06:12 |
Le 11-févr.-08, à 17:58, Mirek Dobsicek a écrit : > >> "But thanks to that crashing, *Church thesis remains consistent*. I >> would just say "An existence of a universal language is not ruled >> out". >> >> >> >> I am ok with you. Consistent (in math) means basically "not rule out". >> "Formally consistent" means "not formally ruled out", or "not >> refutable". >> >> That is: >> >> "Consistent(p") is the same as "~ Provable(~ p)" " ~" = negation >> >> like "Provable(p)" is equivalent with "~ Consistent( ~ p)" > > All right... > > > Now, let me just rephrase few points of the key post one more time. I > will try to be picky with wording. Points which are not mentioned - I > fill comfortable with. > > 1\ There is no language/algorithm/procedure/machine which can > describe/execute all total computable functions. I guess (and have some confirmation below) that you mean here that there is no language/algorithm/procedure/machine which can describe/execute all total computable functions AND ONLY TOTAL COMPUTABLE FUNCTIONS. Right? Otherwise you would be saying that Church Thesis is false.
> 2\ There exists non-empty set of all machine computable functions > (inevitably includes both total and strict partial functions). Let us > call this set MC (machine computable). OK. (this confirms what I say above).
> 3\ Church himself *defined* the set of > so-far-known-computable-functions > as those computable by lambda calculus.
Well, for Church, Lambda does define the set of all computable (total and partial) functions, not just the so-far-known-comp-functions. > 4\ What we use to call a *Church thesis* today says, that MC is in fact > equal to the set of functions computable by lambda calculus. Yes.
> 5\ Church thesis is refutable.
Yes. (It is enough to give a function which we can compute, but which would not be Turing or Lambda computable.
> > > >> * * * >> >> Something else: >> >> to us verify MM = SII(SII) does crash the system: >> >> SII(SII) = I(SII)(I(SII)) = SII(SII) = I(SII)(I(SII)) = SII(SII) = >> I(SII)(I(SII)) = SII(SII) = I(SII)(I(SII)) = SII(SII) = ... >> (crashing). >> > > Working with SK combinators, I had a bit problems with omitted > parenthesis.
Just avoid writing the left parentheses. SII(SII) is really put for (((SI)I)((SI)I)) which is conceptually clearer, but practically unreadable.
> Also it was not clear to me what is the meaning of eg. > (SKK) since S is expected to take three arguments.
OK, my fault. It means the construct is stable and waiting for more argument. so SKK, without any added inputs remains SKK and constitutes a combinator described as being SKK. Such construct are useful for making data structures for example. Such construct are called normal forms. The idea is that the 2-variables function Lx Ly (x + y) when applied on just one argument, 5, say, gives a new function Ly (5 + y), which is a 1-variable function adding 5 to its input. It is the way Schoenfinkel managed to have only function of one argument.
> What helped me was > the unlambda language (http://www.madore.org/~david/programs/unlambda/) At first sight this is just lambda calculus (and thus combinator) with awkward notation .... > > So here is my crashing sequence (yep, yours is shorter) (I don't expand > the I term to keep it short)
Good idea. "I" is really a "macro" for "SKK". I hope everybody see that SKKx = x for any x. SKKx = Kx(Kx) = x. (cf Sabc = ac(bc); Kab = a) > SII(SI(S(KI)I))
Hmmm... ok, that's working, but S(KI)I is really I, and you could have make something simpler ...
> > a reference implementation in unlambda: > ```sii``si``s`kii > the ` stands for 'apply' operation, aka left parenthesis. You really need spectacles here ... > > with a small modification > ```sii``si``s`k.ui > we can achieve the computer to print uuuuuuuuuuuuu.... in an endless > loop. .u is a special function with a side effect of printing the > character u. You are not supposed to use anything but K and S (or anything you have already define with K and S) ... I'm not completely sure what you mean by k.ui (the dot) ... hmm... I see, it prints, well ok then.
I will try to comment your "UDA paper" post asap. Best, Bruno http://iridia.ulb.ac.be/~marchal/ |
| Re: UDA paper |
Bruno Marchal |
14/02/08 03:00 |
Hi Mirek, Le 12-févr.-08, à 23:20, Mirek Dobsicek a écrit :
> > Hi Bruno, > >> The UDA, in english, can be found here: >> */The Origin of Physical Laws and Sensations/*, (Invited Talk SANE >> 2004). >> Click on that title, or copy the following in your browser: >> http://iridia.ulb.ac.be/~marchal/publications/ >> SANE2004MARCHALAbstract.html >> (if you study it I would suggest you print the slider too, so that you >> could perhaps tell me which step you would find hard to go through >> ....). > > I have started reading this paper. Just a quick question. > > At the first step of UDA it seems you restrict yourself to classical > bits. That is fine. I can imagine that somebody deliberately read and > cut my running computer so that the computer goes on with its job after > being 'reincarnated' in Helsinky. Even the substitution level is more > or > less clear. Noise on transistors is definitely not important. OK.
> > > However, at the third step you mention quantum mechanics.
This is weird. I read it twice and don't see where I would mention quantum mechanics ? Please note that the 1-indeterminacy I am talking about in the third step is really a pure classical indeterminacy. It arises from the fact that my classical state is duplicable, and then I cannot predict which *experience* I will *feel* after a self-duplication: mainly Washington OR Moscow (or Sidney *or* Beijing), ...
> It is not > clear to me how would you classically teleport my quantum computer. > What > are the read & cut operations? This is a very different question. I just cannot classically teleport a quantum computer. And the UDA is supposed to already justify why we cannot teleport classicaly any "piece of matter". The rough reason is that matter simply not exist, and what we called "matter" is just a rough description of what is observable and that emerges, a priori by comp, from infinities of infinite computations. But this is part of the conclusion of the UD Argument. Few people seems to realize that the violation of Bell's inequality or the non cloning theorem is an easy consequence of the comp hyp. I think people does not realize this because they are not used to take the difference between first and third person points of view seriously enough. This is perhaps a consequence of 1500 years of Aristotelian brainwashing I'm afraid. Or they are just confused by the fact that scientific argument cannot make reference to personal feelings or points of view, although of course scientific argument can *bear* upon such personal experiences, through definitions, axioms, discourses, etc.
> > Yes, there exists a classical Turing machine which can simulate my > quantum computer,
Yes, but only by running those infinities of infinite (classical) computations (up to some hard to define equivalence relation: the lobian interview is all what I found to tackle this, and this is a lot because it has to eventually distinguish between loop gravity and superstring theory or whatever the correct third person description is correct).
> but I am not giving the running simulator to you. I > don't have it. Ah but this is not true. Of course you can give me the running simulator ... in case you do accept Church Thesis. The running simulator *is* the UD, which exists by Church Thesis. The UD, globally does "run" all relative states from which, from your first person (plural) point of view, quantum computation emerges (if both comp and the quantum hyp are correct). It does it an infinity of "times" (in Arithmetical Platonia). Although an unknown quantum state is not clonable, it is "preparable" in infinities of examplars. We cannot recognize it in any third person way, yet, we cannot not "recognize" it, albeit implicitly, when we are "living" it. Of course this is step seven ...
> > Please, make a short clarification about your framework. I might be > just > misinterpreting you. My hypothesis is that we are Turing-emulable, at some level of self-description. My conclusion is that whatever the "Universe" is, it cannot be Turing emulable, and in fine the physical laws emerges from machine theology (say). This makes the comp hyp testable: just derive some comp-physics, and compare it to empirical physics.
> What is the page reference to Gruska's book? In the footnote 9 of the SANE paper I am just alluding to the non-cloning theorem which, if I remùamber correctly, is well proved in two manners in the book of Gruska. Just look at Gruska's book index on "non cloning". I don't have under my hands my exemplar right now. I hope this helps a bit. The key point: I am not mentioning or using QM at all in the UDA, except for illustrating how the comp-physics, with its many histories (computation from first person perspectives) and non cloning phenomena is already similar to the "empirical physics".
I hope this helps, Best, Bruno http://iridia.ulb.ac.be/~marchal/ |
| Re: UDA paper |
Bruno Marchal |
20/02/08 06:20 |
Hi Mirek, I guess you are busy. I would just like to insist that when I say (14-febr.-08): > Please note that the 1-indeterminacy I am talking about in the third > step is really a pure classical indeterminacy. It arises from the fact > that my classical state is duplicable, and then I cannot predict which > *experience* I will *feel* after a self-duplication: mainly Washington > OR Moscow (or Sidney *or* Beijing), ...
This is really a key point, if not *the* key point. I think it is almost trivial, but sometimes some people have a problem with this. In that case it helps to imagine the same experiment done with some inference inductive machine in place of a human or "you", and this in an iterated self-duplication. In that case the result amount to saying that no robot, when duplicated iteratively (in Washington and Moscow, say) can predict its future sequence of results of first person self-localization. This becomes equivalent with the fact that most finite bit-strings, like WMMMWWMWWWM ... are not compressible. Someone told me (out-of-line) that he *can* predict with certainty his future in that situation: for example he can predict WWWWWWWWW..., but this means he is not taking into account the saying of the other reconstituted people, which, *assuming comp* are genuine "descendant" of the "original". Those people will acknowledge that their "prediction with certainty" was false, and they have the same right and reason to be taken seriously, again when we *assume* the comp hypothesis.
Have you a problem with this? I think most on this list grasp this point, but don't hesitate to tell me if you don't. Without a clear understanding of what happens here we can't really proceed ... (nor can we grasp Everett formulation of QM I could argue ...). Bruno http://iridia.ulb.ac.be/~marchal/ |
| Re: UDA paper |
Mirek Dobsicek |
20/02/08 07:24 |
Hi Bruno, yes, I am now a bit busy. Lecturing, seminars,.. wedding planning :-) I am somewhere in the middle your paper. Regarding the very point of the described 1-indeterminancy, I have no problem there at all. Anyone who ever called a fork() unix function (read, cut, duplicate) followed by an execve(read-destination-name-from-keyboard()) function, should not have a problem here. A program, even with considerably good self-referential skills, has no chance to know whether I will enter Warsaw or Moscow on the keyboard. As I have said, I have not finished reading the paper yet. But sometime I have a problem with a bit of feeling of circularity of arguments, or described in better words, given assumptions A={..}, conjectures B={...} are true, where Bs feels like rephrased As, and therefore Bs are trivially true. No disrespect here! It just how do I feel now. Bs are overwhelming, but As are pretty strong assumptions, so Bs are not surprising anymore, yet an hour later Bs are overwhelming again. Best, Mirek |
|
Lennart Nilsson |
20/02/08 08:15 |
-----Ursprungligt meddelande----- Fr=E5n: everyth...@googlegroups.com [mailto:everyth...@googlegroups.com] F=F6r Bruno Marchal Skickat: den 20 februari 2008 15:21 Till: everyth...@googlegroups.com =C4mne: Re: UDA paper > It arises from the fact
> that my classical state is duplicable... And of course your quantumstate is not...
So your argument that the duplication can be said to be on any level, including a whole universe if need be, is not an airproof argument? LN |
| Re: |
Bruno Marchal |
21/02/08 03:09 |
Le 20-févr.-08, à 17:15, Lennart Nilsson a écrit : > > > > -----Ursprungligt meddelande----- > Fr=E5n: everyth...@googlegroups.com > [mailto:everyth...@googlegroups.com] F=F6r Bruno Marchal > Skickat: den 20 februari 2008 15:21 > Till: everyth...@googlegroups.com > =C4mne: Re: UDA paper > > >> It arises from the fact >> that my classical state is duplicable... > > > And of course your quantumstate is not... > > So your argument that the duplication can be said to be on any level, Perhaps. This is ambiguous. The comp hyp *assumes* that there is a level of description of me such that "I survive" (= I experience no change) when my "body" (= the thing on which my consciousness supervenes on) is piece by piece substituted at that level.
> including a whole universe if need be, ... just In the (not plausible but logically conceivable) case my "body" (see above) *is* the whole physical universe. In *that* case, by definition, the comp hyp asks the whole physical universe to be turing-emulable. This is akin to a form of cosmic solipsism. If this is true Schmidhuber and constructive physicists would be (almost trivially) right. But, they would still be strategically wrong by *assuming* this at the start, wrong with respect to the mind-body problem (or with respect to the problem of the relation between first and third person discourses). The UD Argument is supposed to show that the comp hypothesis forces us to derive the physics from a "probability" bearing on the machine's personal (hopefully plural first person) comp histories, if the UD A is correct.
> is not an airproof argument? What argument? I think you are only discussing the meaning of the starting assumption here. Have you grasp the whole 8-steps argument? If I'm wrong or unclear just tell me where and let us discuss where the precise problems are. Please keep in mind that I am open to the idea that the physics extracted from comp is incompatible with the empirical physics making comp not sustain by empirical evidences. Perhaps you could also tell me what is your opinion on Everett or Deutsch. People who dislikes Everett's work could hardly appreciate mine.
Bruno http://iridia.ulb.ac.be/~marchal/
|
| assumptions |
Lennart Nilsson |
21/02/08 06:01 |
I think you are only discussing the meaning of the starting assumption here. Have you grasp the whole 8-steps argument? If I'm wrong or unclear just tell me where and let us discuss where the precise problems are. Please keep in mind that I am open to the idea that the physics extracted from comp is incompatible with the empirical physics making comp not sustain by empirical evidences. Perhaps you could also tell me what is your opinion on Everett or Deutsch. People who dislikes Everett's work could hardly appreciate mine.
Bruno http://iridia.ulb.ac.be/~marchal/
Oh, I am very much proEverett and proDeutsch and, I might add proStandish (having translated his book Theory of Nothing into swedish). And you are right of course, it was your assumptions I questioned...
LN |
| Re: UDA paper |
Bruno Marchal |
22/02/08 06:32 |
Hi Mirek, Le 20-févr.-08, à 16:24, Mirek Dobsicek a écrit : > yes, I am now a bit busy. Lecturing, seminars,.. wedding planning :-)
Waouh! Congratulation! I'm happy for you.
> > I am somewhere in the middle your paper. Regarding the very point of > the > described 1-indeterminancy, I have no problem there at all. Anyone who > ever called a fork() unix function (read, cut, duplicate) followed by > an > execve(read-destination-name-from-keyboard()) function, should not have > a problem here. A program, even with considerably good self-referential > skills, has no chance to know whether I will enter Warsaw or Moscow on > the keyboard. Nice to hear that.
> > As I have said, I have not finished reading the paper yet. But sometime > I have a problem with a bit of feeling of circularity of arguments, or > described in better words, given assumptions A={..}, conjectures > B={...} > are true, where Bs feels like rephrased As, and therefore Bs are > trivially true. No disrespect here!
Not at all. I appreciate. Sometimes I have to explain lengthily that my contribution is modest (even if a bit radical). And not so original if you take into account that the comp-like (platonist) conception of reality has been defended by many greek intellectuals during a millenium, before being banished or murdered like Hypatia.
> > It just how do I feel now. Bs are overwhelming, but As are pretty > strong > assumptions,
Well, at the same time, not so much. Most scientist believe (not always consciously) in comp, and at the same time in some notion of *primary* or *primitive* matter. They feel dizzy when they begin to understand the incompatibility between comp and (weak) materialism. Sometimes comp or mechanist philosophy is used by materialist to put the mind/body problem under the rug, which explains why they dislike my work. Sometimes it is even just political: they believe I am attacking Marx or Lenine ... The main contribution I have done (I think) is in the illustration that by making comp sufficiently precise, some of the weirder aspect are testable. Here the results shock many people among those who does not know the current "interpretation" problem of QM. When I talked on many-worlds in the seventies, it was enough to be put in the crackpot category. I am sure many on the list have lived similar things. > so Bs are not surprising anymore, yet an hour later Bs are > overwhelming again.
Yes it is like that. Few people realise that comp *is* a very strong assumption, even just Church thesis is already very strong, and has many counter-intuitive consequences. This is not very well know too.
> > Best, > Mirek I wish you the best to you and to your girlfriend. She is lucky, you look serious. I hope your (future) wife will not trow the books she offered to you through the windows, like it happened to a friend of mine (and she threw the computer with!). I reassure you: I think that was exceptional! Presently I am not so lucky because I have been break in yesterday, and my home computer has been stolen with all the attached devices including the main backup disk. I will have to rewrite hundreds of unfinished papers... . Sorry to bother you with that, actually.
Have a super nice wedding, and take all the time you need to read the few papers I have finished, Bruno
http://iridia.ulb.ac.be/~marchal/ |
| Re: assumptions |
Bruno Marchal |
22/02/08 06:40 |
Le 21-févr.-08, à 15:01, Lennart Nilsson a écrit : > Oh, I am very much proEverett and proDeutsch and, I might add > proStandish > (having translated his book Theory of Nothing into swedish). And you > are > right of course, it was your assumptions I questioned... OK, no problem then, Have a good day, Bruno http://iridia.ulb.ac.be/~marchal/
|
| Re: UDA paper |
Russell Standish |
23/02/08 00:43 |
On Fri, Feb 22, 2008 at 03:32:28PM +0100, Bruno Marchal wrote: > Presently I am not so lucky because I have been break > in yesterday, and my home computer has been stolen with all the > attached devices including the main backup disk. I will have to rewrite > hundreds of unfinished papers... . Sorry to bother you with that, > actually. > > Have a super nice wedding, and take all the time you need to read the > few papers I have finished, > > > Bruno > What a drag! Its one thing getting your computer stolen, but making off with the backup too is really cruel. I hope you can recover some things... Cheers -- ---------------------------------------------------------------------------- A/Prof Russell Standish Phone 0425 253119 (mobile) Mathematics UNSW SYDNEY 2052 hpc...@hpcoders.com.au Australia http://www.hpcoders.com.au ----------------------------------------------------------------------------
|
| Re: UDA paper |
Tom Caylor |
23/02/08 11:27 |
On Feb 22, 7:32 am, Bruno Marchal <marc...@ulb.ac.be> wrote:
> I reassure you: I think that
> was exceptional! Presently I am not so lucky because I have been break
> in yesterday, and my home computer has been stolen with all the
> attached devices including the main backup disk. I will have to rewrite
> hundreds of unfinished papers... . Sorry to bother you with that,
> actually.
>
> Have a super nice wedding, and take all the time you need to read the
> few papers I have finished,
>
> Bruno
I'm also sorry to hear that, Bruno. Hundreds of unfinished papers!
Whoa! Good and bad. Some mourning to do. Mourning is good. I just
taught a lesson this morning on the Lamentations (in the old testament
poetic writings). First step of restoration ("getting back what was
lost") is stepping back and taking a deep breath and accepting where
we are now, accepting present reality. Then it is remembering "the
precious things". In your case it could be the precious kernels of
gold in all your work that was lost. Mourning the loss, but
celebrating that it was a loss that contained something that was very
valuable, that you actually had something valuable in all of that, and
no one can ever take that from you. You will probably find that the
kernel has something to do with relationships: with other persons,
with Truth. Third is to realize that we cannot restore our life/
strength/soul ourselves (this is where most of us get stuck a lot).
Counter to the modern/reductionist approach, restoring what was lost
does not depend on simply re-writing papers. It has to come from the
Source of all. This can be an opportunity to remember (or find out in
a new way) what really matters. A side note: it is interesting that
in your pursuit of truth there is a lot that remains unfinished, it
seems even more than what has been finished. I think this is a
comment/tribute to your theories, that truth is found on the forever-
unfinished border between the known and unknown. It is a journey!
Take courage. And thanks for your contributions to our thinking.
Tom
|
| Re: UDA paper |
Mirek Dobsicek |
24/02/08 06:21 |
Hi Bruno! > I wish you the best to you and to your girlfriend. Thank you very much. I appreciate your wish. > offered to you through the windows, like it happened to a friend of > mine (and she threw the computer with!). I reassure you: I think that > was exceptional! Presently I am not so lucky because I have been break > in yesterday, and my home computer has been stolen with all the > attached devices including the main backup disk. I will have to rewrite > hundreds of unfinished papers... . Sorry to bother you with that, > actually. Ohh, that is not good. I am sorry for you. Hopefully you can recover at least parts of the most important things. Sincerely, Mirek
|
| Re: UDA paper |
JohnM |
25/02/08 06:02 |
Bruno,
my deepest sympathy to your computer-loss!!
it comes up frequently in my nightmares to have a similar 'catastrophe' on my own and the desperation wakes me that I cannot recover my 'recent past' before I have recovered it, like the lost eyeglasses what you cannot look for until you found it.
I had one such deluge-loss of data, when the very first virus was loanched in the 90s, I fell into it right before my wife warned me about such danger as told in the actual News on TV.
I lost all my graphics - irrecoverably. Of course this is no consolation, but you may have most of the material on other computers and on paper.
A terrible job and please, forgive my flippant remark on the recovery of your 'unfinished' texts:
Have a "Normal Hauptman" read the recovered texts...-- I try to explain:
The Austria-Hungary War Ministry had a position of a "Normal Hauptman" ('average lieutenant') who's job was to READ all ordinances planned to be sent to the troops. If he understood them and could tell what they said, it was issued.
I do not volunteer for this position with you: find a smarter one...
Best wishes
John (too normal)
|
| Re: UDA paper |
Bruno Marchal |
25/02/08 06:19 |
Russell, Tom, Barry, Mirek, Thank you very much for your kind posts. Actually I am amazed by the quantity of backup I have done, and so it looks like I have recovered all the main "professional" things, including most non finished papers (and then I agree with Tom's wise remarks which make me not so sure that I am glad with recovering those unfinished works ... 'cause I have no reason to not finish them now, or I am must search some reason ...). It looks I have only loss the leisure private documents (recent pictures, private mails), and many movies I have done, but not all, of transformations in the complex plane illustrating universal computability in the complex plane. Of course it is never funny when unknown people put some mess in your stuff, but it is not so grave, and those first person happenings are relative. To thank you I send you one of those (remaining) movie at your personal address. If Wei Dai agree, I could send it online: it is a 1,5 Mega QuickTime document attachment. I guess it is a bit too big. Some day I will put them on my web page. It does illustrate some points. The problem is that my all complex plane software does no more run on current computers. Best, Bruno http://iridia.ulb.ac.be/~marchal/ |
| Re: UDA paper |
Bruno Marchal |
25/02/08 07:16 |
Hi John, Le 25-févr.-08, à 15:02, John Mikes a écrit : > Bruno, > my deepest sympathy to your computer-loss!! > it comes up frequently in my nightmares to have a similar > 'catastrophe' on my own and the desperation wakes me that I cannot > recover my 'recent past' before I have recovered it, It sounds awful! > like the lost eyeglasses what you cannot look for until you found it. I have often a similar problem: without my eyeglasses I cannot see them ... (unless they are infinitely far from me!)
> I had one such deluge-loss of data, when the very first virus was > loanched in the 90s, I fell into it right before my wife warned > me about such danger as told in the actual News on TV. > I lost all my graphics - irrecoverably. What a pity. Someone told me the story of an employee who has encoded typed texts on a computer for four years without doing backup and then it crashed! We often hear stories like that. > Of course this is no consolation, but you may have most of the > material on other computers and on paper. Yes. I have just get a moment of panic for some important document of the year 2002-2004, where I used a PC, but I found eventually that I did have make a double backup home/university of that PC although I hardly remember it. Before and after those years I was and still am using a MAC. Actually I was very glad with that PC except when I begun to be attacked by viruses which have make me decide to come back on Mac. I did never succeed to eliminate the viruses on that PC. Fortunately, those viruses were not of the destructive type, only terribly annoying, slowing down the machine a lot, and distracting with many irrelevant messages. > A terrible job and please, forgive my flippant remark on the recovery > of your 'unfinished' texts: > Have a "Normal Hauptman" read the recovered texts...-- I try to > explain: > The Austria-Hungary War Ministry had a position of a "Normal Hauptman" > ('average lieutenant') who's job was to READ all ordinances planned to > be sent to the troops. If he understood them and could tell what they > said, it was issued. > I do not volunteer for this position with you: find a smarter one...
Thanks for the idea. Best regards, Bruno PS If you want I can send you the 1,5 mega complex-UD-movie, tell me out-of-line (I don't want to take the risk of crashing your computer!). In fact I am not sure it can go through because it could be that there is a limitation access on the university band.
> > John (too normal) > On Sun, Feb 24, 2008 at 9:21 AM, Mirek Dobsicek <m.dob...@gmail.com> > wrote: >> >> Hi Bruno! >> >> > I wish you the best to you and to your girlfriend. >> >> Thank you very much. I appreciate your wish. >> >> > offered to you through the windows, like it happened to a friend of >> > mine (and she threw the computer with!). I reassure you: I think >> that >> > was exceptional! Presently I am not so lucky because I have been >> break >> > in yesterday, and my home computer has been stolen with all the >> > attached devices including the main backup disk. I will have to >> rewrite >> > hundreds of unfinished papers... . Sorry to bother you with that, >> > actually. >> >> Ohh, that is not good. I am sorry for you. Hopefully you can recover >> at >> least parts of the most important things. >> >> Sincerely, >> Mirek >> >> >>
http://iridia.ulb.ac.be/~marchal/ |
| Re: assumptions |
Bruno Marchal |
27/02/08 06:32 |
Le 21-févr.-08, à 15:01, Lennart Nilsson a écrit :
> Oh, I am very much proEverett and proDeutsch and, I might add
> proStandish > (having translated his book Theory of Nothing into swedish). And you > are > right of course, it was your assumptions I questioned... OK, and I said : "No problem then", because (as a logician perhaps) I seldomly discuss assumptions.
But still, although I agree with Mirek that comp (as I define it, i.e. CT + "yes doctor") is a very strong hypothesis, the negation of it is also quite strong, if not stronger. For believing in non-comp, you have to believe in actual physical infinities and/or in genuine third person indeterminacy (like copenhagen and anti-everettist). You could perhaps elaborate abit, if you have the time. Bruno
http://iridia.ulb.ac.be/~marchal/
|
| Re: UDA paper |
Wei Dai |
27/02/08 11:46 |
Bruno Marchal wrote: > If Wei Dai agree, I could send it online: it is a 1,5 Mega QuickTime > document attachment. I guess it is a bit too big. Some day I will put > them on my web page. It does illustrate some points. The problem is > that my all complex plane software does no more run on current > computers. Bruno, you can upload it to http://groups.google.com/group/everything-list/files.
|
| Re: UDA paper |
Bruno Marchal |
29/02/08 06:02 |
Hi Wei, I have not succeeded to upload the movie, nor do I have seen files which I heard should have been already uploaded by people on the list. The system complains that I am not a member of the list. I will try again Monday, because it looks like the discussion are not currently available too, so the problem is perhaps with the Googlegroups. But if that works it is of course the good idea, thanks, Bruno http://iridia.ulb.ac.be/~marchal/
------------------------- Le 27-févr.-08, à 20:46, Wei Dai a écrit : http://iridia.ulb.ac.be/~marchal/ |
| Re: UDA paper |
Torgny Tholerus |
29/02/08 06:25 |
Bruno Marchal skrev: > Hi Wei, > > I have not succeeded to upload the movie, nor do I have seen files > which I heard should have been already uploaded by people on the list. > The system complains that I am not a member of the list. > I will try again Monday, because it looks like the discussion are not > currently available too, so the problem is perhaps with the > Googlegroups. > > But if that works it is of course the good idea, thanks, > >
I have just tested to upload a file to the group (PofSTorgny1.doc). You can try to see if you can see that file. (You have to log in to Google groups first.) -- Torgny Tholerus
|
| Re: UDA paper |
Bruno Marchal |
03/03/08 02:51 |
|
| Re: UDA paper |
Bruno Marchal |
04/03/08 05:45 |
Hi Torgny, Le 29-févr.-08, à 15:25, Torgny Tholerus a écrit : > I see (and did print) your file. I have put the movie there, in two version but I cannot retrieve it. With the first I get the code, and with the other (the one with ".mpeg") I get the QuickTime logo with an interrogation mark. If you or someone can see the movie from there, just tell me. Best, Bruno http://iridia.ulb.ac.be/~marchal/ |
| Re: UDA paper |
Torgny Tholerus |
04/03/08 09:04 |
Bruno Marchal skrev: > Hi Torgny, > > Le 29-févr.-08, à 15:25, Torgny Tholerus a écrit : > > >>
>> I have just tested to upload a file to the group (PofSTorgny1.doc). >> You >> can try to see if you can see that file. (You have to log in to Google >> groups first.) >> > > I see (and did print) your file. I have put the movie there, in two > version but I cannot retrieve it. With the first I get the code, and > with the other (the one with ".mpeg") I get the QuickTime logo with an > interrogation mark. If you or someone can see the movie from there, > just tell me. > > I have not succeeded to view your movie. I have downloaded your files to my computer. But it seems as if your files are corrupted in some way. I have tried three different movie players (Windows Media Player, RealPlayer, and QuickTime), but no one was able to recognize your files. -- Torgny
|