> Charlie-Boo <shymath...@gmail.com> writes: > > On Nov 5, 4:45 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote: > >> Charlie-Boo <shymath...@gmail.com> writes: > >> > On Nov 5, 4:02 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote: > >> >> Charlie-Boo <shymath...@gmail.com> writes: > >> >> > On Nov 5, 2:09 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote: > >> >> >> Charlie-Boo <shymath...@gmail.com> writes: > >> >> >> > On Nov 5, 10:29 am, James Burns <burns...@osu.edu> wrote: > >> >> >> >> Charlie-Boo wrote: > >> >> >> >> > Is there a two-place relation R such that:
> >> >> >> >> > 1. If x is an element of y then there exists a z such that R(y,z).
> >> >> >> >> > 2. If R(x,y) then y is an element of x.
> >> >> >> >> > 3. If R(x,y) and R(x,z) then y=z.
> >> >> >> >> > What should it be called?
> >> >> >> >> Would you mind sharing why the question is interesting?
> >> >> >> > A lot of people write about it.
> >> >> >> How about some context then? Who writes about it?
> >> >> > Does the name ``Godel" ring a bell?
> >> >> Yes, but Goedel didn't prove his famous theorems in set theory,
> >> > Why would they be theorems if he didn't prove them?
> > > I didn't say that he didn't prove them.
> > You said, "Goedel didn't prove his famous theorems"
> Right. You've quite a skill
*blush*
> I thought you were going on about the incompleteness > theorems, but you were talking about the independence of CH, so never > mind what I said above.
Trim it?
> >> >> so I still haven't a clue what you're going on about.
> >> > Godel proved theorems in set theory as well as in logic.
> >> Yes, that's true. But where did he assume the existence of this > >> relation R that you're going on about?
> >> Don't be coy. Just spell it out.
> > The choice function referred to in the Axiom of Choice.
> Sorry, still not clear on what you mean. The axiom of choice does not > involve a relation R that you described.
Ok. # 1 = AOC doesn’t involve my R.
> The relation R that you > described would be more like what one would see in an axiom of choice > for classes.
Ok. # 2 = AOC involves my R when talking about classes.
But # 1 => ~(# 2).
> Tell you what. Why don't you write down the axiom of choice and point > out where it involves such an R?
Let aoc() be the choice function. Then aoc(x)=y iff R(x,y).
> >> (1) has a free variable x. It's not clear what you mean by (1).
> > I would say that x is universally quantified.
> Okay, so you mean to ask: is there a relation R such that
> The antecedent is always false and hence the conditional is true, > regardless of what R is -- unless I'm making some silly error as I > toss this off. So, I doubt this is really what you meant after all.
> >> You think I'll purchase and read this book to figure out what you're > >> talking about? I have a better idea. You can just explicitly state > >> which relation you mean (and, perhaps, where Goedel introduces this > >> relation).
> > Godel proved that the Axiom of Choice is consistent with ZF.
> -- > Jesse F. Hughes > "You do know that after the get done with [outlawing] cigarettes, > they're gonna come after guns, right?" > -- AM talk radio host Mike Gallagher- Hide quoted text -
> Tim Little <t...@little-possums.net> writes: > > On 2009-11-05, Jesse F. Hughes <je...@phiwumbda.org> wrote: > >> Right. You've quite a skill
> > Heh. Heh.
> >> (Ax)(Ay)( x e y -> (Ez)( R(y,z) ) )?
> >> This is equivalent to
> >> (Ay)( ( (Ax) x e y ) -> (Ez)( R(y,z) ) ).
> >> The antecedent is always false and hence the conditional is true, > >> regardless of what R is -- unless I'm making some silly error as I > >> toss this off.
> > Yes, unfortunately. Quantifiers do not distribute over implication > > like that. The first statement asserts something about all nonempty > > sets y, while the second asserts something about a universal set y.
> D'oh! Eh, that's what I get for trying to toss off a reply while I'm > heading out the door.
> -- > Jesse F. Hughes > "Now 'pure math' makes sense as well as clearly it's a peacock game, > where some of you see it as a way to show you as being highly > intelligent and thus more desirable to women." -- James S. Harris- Hide quoted text -
Charlie-Boo <shymath...@gmail.com> writes: > On Nov 5, 7:31 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote: >> Tim Little <t...@little-possums.net> writes: >> > On 2009-11-05, Jesse F. Hughes <je...@phiwumbda.org> wrote: >> >> Right. You've quite a skill
>> > Heh. Heh.
>> >> (Ax)(Ay)( x e y -> (Ez)( R(y,z) ) )?
>> >> This is equivalent to
>> >> (Ay)( ( (Ax) x e y ) -> (Ez)( R(y,z) ) ).
>> >> The antecedent is always false and hence the conditional is true, >> >> regardless of what R is -- unless I'm making some silly error as I >> >> toss this off.
>> > Yes, unfortunately. Quantifiers do not distribute over implication >> > like that. The first statement asserts something about all nonempty >> > sets y, while the second asserts something about a universal set y.
> > D'oh! Eh, that's what I get for trying to toss off a reply while > I'm > > heading out the door.
> I thought haste makes waste, not stupidity.
It was a stupid mistake. I'm sure you've never made a silly blunder yourself, so obviously you have the right to mock me.
-- "It's one of the easiest tickets to true fame--not this silly stuff where people cheer you for a few years and then forget about you--but the kind of fame where school kids have to read your biography and do reports on you." -- Another reason to support James S. Harris.
Charlie-Boo <shymath...@gmail.com> writes: >> Sorry, still not clear on what you mean. The axiom of choice does not >> involve a relation R that you described.
> Ok. # 1 = AOC doesn’t involve my R.
>> The relation R that you >> described would be more like what one would see in an axiom of choice >> for classes.
> Ok. # 2 = AOC involves my R when talking about classes.
> But # 1 => ~(# 2).
You're talking nonsense. The axiom of choice refers to a particular (equivalence class of) axiom(s). It is an axiom about sets. The class-based axiom of choice that I mentioned (which I've never seen in the literature) is a different axiom.
Thus #1 does not entail ~(#2).
-- "There's lots of things in this old world to take a poor boy down. If you leave them be, you can save yourself some pain. You don't have to live in fear, but you best have some respect, For rattlesnakes, painted ladies and cocaine." -- Bob Childers
Charlie-Boo <shymath...@gmail.com> writes: >> Tell you what. Why don't you write down the axiom of choice and point >> out where it involves such an R?
> Let aoc() be the choice function. Then aoc(x)=y iff R(x,y).
Wow. What an utter failure to write down the axiom of choice. Want to try again?
You speak, after all, as if there is a single choice function. Tain't so. -- Jesse F. Hughes "This Trojan appears to utilize a function of the Windows Media DRM designed to enable license delivery scenarios as part of a social engineering attack." -- MS candidly explains the role of DRM licenses
> On Thu, 5 Nov 2009, Charlie-Boo wrote: > > Is there a two-place relation R such that:
> Yes.
> > 1. If x is an element of y then there exists a z such that R(y,z).
> x in y ==> some z with R(y,z)
> > 2. If R(x,y) then y is an element of x.
> R(x,y) ==> y in x
> R(x,y) ==> y in x ==> some z with R(x,z) ==> some z with z in x
> > 3. If R(x,y) and R(x,z) then y=z.
> R is a function
> > What should it be called?
> Choice function.
> What's the domain and codomain of R?
Anything as long as R satisfies 1-3.
The idea is to,
1. Formalize the Axiom of Choice using well-known (well-understood) primitives: Predicate Calculus.
2. Use logic to develop simpler requirements that if impossible make AOC impossible. This simplifies the question of whether AOC is true or not. (1) is an example.
3. Apply incompleteness proofs in other domains e.g. Computability to this formalization.
Define,
YES(x,y) iff Turing Machine x halts yes on input y. SE(x,y) iff y is an element of x.
M defines r.e. set YES(M,x) and (general) set SE(M,x).
There is no M that defines an r.e. set ~YES(x,x). There is no M that defines a (general) set ~SE(x,x).
Thus we show there is no set of sets that contain themselves.
With a little bit of logic we can likewise say there is no r.e. set (exists y)~YES(x,y) and similarly with other wffs (theorems) of Computability.
If we substitute YES for SE in the definition of AOC or its necessary conditions 1-3, we can very directly manipulate that wff as referring to Turing Machines and e.g. appeal to known theorems. Then we apply that same manipulation to SE.
BTW: If there is an R that meets (1) it doesn’t necessarily meet (2) or (3). However, does the existence of such an R mean there is some R that meets (1) and (2)? (1) and (3)? Which of the 8 subsets of 1-3 are equivalent to which others in this sense? This would even more directly reduce AOC to simpler questions.
> Charlie-Boo <shymath...@gmail.com> writes: > > On Nov 5, 7:31 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote: > >> Tim Little <t...@little-possums.net> writes: > >> > On 2009-11-05, Jesse F. Hughes <je...@phiwumbda.org> wrote: > >> >> Right. You've quite a skill
> >> > Heh. Heh.
> >> >> (Ax)(Ay)( x e y -> (Ez)( R(y,z) ) )?
> >> >> This is equivalent to
> >> >> (Ay)( ( (Ax) x e y ) -> (Ez)( R(y,z) ) ).
> >> >> The antecedent is always false and hence the conditional is true, > >> >> regardless of what R is -- unless I'm making some silly error as I > >> >> toss this off.
> >> > Yes, unfortunately. Quantifiers do not distribute over implication > >> > like that. The first statement asserts something about all nonempty > >> > sets y, while the second asserts something about a universal set y.
> > > D'oh! Eh, that's what I get for trying to toss off a reply while > > I'm > > > heading out the door.
> > I thought haste makes waste, not stupidity.
> It was a stupid mistake.
Good. Thanks.
> I'm sure you've never made a silly blunder > yourself, so obviously you have the right to mock me.
Hmmm . . . Doesn't everyone have the right to mock anyone? Or at least the same rights?
My point is that attributing a mistake to haste leaves something to be desired. (1) Why bother - what's the point? Shouldn't we ALWAYS not judge something someone did by judging something else that they did? So it has no relevance to anything. (2) It is a little suspiocious when someone says they said something due to haste. I would agree that we can attribute it to not taking the time to think about it. Is that what you meant? But that occurs when someone posts an easy problem because they thought it was neat (and very well may be) but didn't then check that it is actually difficult before posting it. But then again, now we are poking a hole in a defense mechanism that needn't be used anyway, so that is a waste. (3) What you did wasn't bad. Bad is using ad hominem. Or worse, defending the use of ad hominem. So it also not worth defending. (4) In general, let's all be big boys and not waste time tending to foolish pride (the root of all evil to many.)
> -- > "It's one of the easiest tickets to true fame--not this silly stuff > where people cheer you for a few years and then forget about you--but > the kind of fame where school kids have to read your biography and do > reports on you." -- Another reason to support James S. Harris.- Hide quoted text -
> Charlie-Boo <shymath...@gmail.com> writes: > >> Sorry, still not clear on what you mean. The axiom of choice does not > >> involve a relation R that you described.
> > Ok. # 1 = AOC doesn’t involve my R.
> >> The relation R that you > >> described would be more like what one would see in an axiom of choice > >> for classes.
> > Ok. # 2 = AOC involves my R when talking about classes.
> > But # 1 => ~(# 2).
> You're talking nonsense. The axiom of choice refers to a particular > (equivalence class of) axiom(s). It is an axiom about sets. The > class-based axiom of choice that I mentioned (which I've never seen in > the literature) is a different axiom.
> Thus #1 does not entail ~(#2).
I guess it depends on your definition of "involves". It is a very broad word to me.
Anyway, how is R about classes and not sets? You may be getting to the point, actually. Classes are for things that are not sets (=relations) so AOC is really about whether R is a set. Russell proved that some things aren't sets and I am trying to apply additional logic to address R being a set or not.
> -- > "There's lots of things in this old world to take a poor boy down. > If you leave them be, you can save yourself some pain. > You don't have to live in fear, but you best have some respect, > For rattlesnakes, painted ladies and cocaine." -- Bob Childers
On Nov 10, 7:53 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> Charlie-Boo <shymath...@gmail.com> writes: > >> Tell you what. Why don't you write down the axiom of choice and point > >> out where it involves such an R?
> > Let aoc() be the choice function. Then aoc(x)=y iff R(x,y).
> Wow. What an utter failure to write down the axiom of choice. Want > to try again?
> You speak, after all, as if there is a single choice function. Tain't > so.
I know. That's good of you to understand the set axioms so well. I like the simpler version. So, do we know that they aren't equivalent?
> -- > Jesse F. Hughes > "This Trojan appears to utilize a function of the Windows Media DRM > designed to enable license delivery scenarios as part of a social > engineering attack." -- MS candidly explains the role of DRM licenses
> > On Nov 5, 3:05 pm, Dan Cass <dc...@sjfc.edu> wrote: > > > I'm answering your question about condition (1) > > only: > > > excerpt....> On Nov 5, 10:02 am, Charlie-Boo > > > > <shymath...@gmail.com> wrote: > > > > > Is there a two-place relation R such that:
> > > > > 1. If x is an element of y then there exists a > > z > > > > such that R(y,z).
> > > > Is there a relation R that meets (1) ? (Or > > something > > > > that is not a > > > > relation.)
> > > > C-B
> > > Suppose R(y,z) means z is an element of y. > > > Then your condition (1) is true, since z = x does > > the job.
> > Ok, but is that R a relation?
> > C-B
> If a relation R is defined as a subset of a product of two sets, then yes.- Hide quoted text -
And is there a set (or superset) of all nonempty sets?
Charlie-Boo <shymath...@gmail.com> writes: > On Nov 10, 7:53 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote: >> Charlie-Boo <shymath...@gmail.com> writes: >> >> Tell you what. Why don't you write down the axiom of choice and point >> >> out where it involves such an R?
>> > Let aoc() be the choice function. Then aoc(x)=y iff R(x,y).
>> Wow. What an utter failure to write down the axiom of choice. Want >> to try again?
>> You speak, after all, as if there is a single choice function. Tain't >> so.
> I know. That's good of you to understand the set axioms so well. I > like the simpler version. So, do we know that they aren't > equivalent?
As I just posted Global AC (your simpler conditions) imply AC, but there is no reason to think that AC implies Global AC as far as I know.
I'd imagine that the proof that countable choice does not imply AC gives a hint as to how one would show Global AC does not imply AC, but I'm not familiar with that argument.
-- Jesse F. Hughes "Yes, I'm one of those arrogant people who tries to be quotable. There is actually at least one person who quotes me often." -- James Harris
Charlie-Boo <shymath...@gmail.com> writes: > On Nov 10, 7:51 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote: >> Charlie-Boo <shymath...@gmail.com> writes: >> >> Sorry, still not clear on what you mean. The axiom of choice does not >> >> involve a relation R that you described.
>> > Ok. # 1 = AOC doesn't involve my R.
>> >> The relation R that you >> >> described would be more like what one would see in an axiom of choice >> >> for classes.
>> > Ok. # 2 = AOC involves my R when talking about classes.
>> > But # 1 => ~(# 2).
>> You're talking nonsense. The axiom of choice refers to a particular >> (equivalence class of) axiom(s). It is an axiom about sets. The >> class-based axiom of choice that I mentioned (which I've never seen in >> the literature) is a different axiom.
>> Thus #1 does not entail ~(#2).
> I guess it depends on your definition of "involves". It is a very > broad word to me.
They are related, but I wouldn't say that AC involves your R.
> Anyway, how is R about classes and not sets? You may be getting to > the point, actually. Classes are for things that are not sets > (=relations) so AOC is really about whether R is a set. Russell > proved that some things aren't sets and I am trying to apply > additional logic to address R being a set or not.
I shouldn't have said that R was about classes per se, but your conditions claim that there is essentially a *global* choice function, that is a choice function for the particular class V. That's not what AC says.
Compare the following:
AC:
For all w, there is an R c w x Uw such that the following three conditions hold:
(Ay in w)( (Ex)( x in y ) -> (Ez) R(y,z) ) (Ay)(Az)( R(y,z) -> z in y ) (Ay)(Az)(Az') ( ( R(y,z) & R(y,z') ) -> z = z' )
Global AC:
There is an R such that the following three conditions hold:
(Ay)( (Ex)( x in y ) -> (Ez) R(y,z) ) (Ay)(Az)( R(y,z) -> z in y ) (Ay)(Az)(Az') ( ( R(y,z) & R(y,z') ) -> z = z' )
Those are two different claims. In Global AC, it is clear that R cannot be a set at all. It must be a proper class of ordered pairs.
Clearly, Global AC implies AC. Suppose w is a set and let R be given as in Global AC. Define
R' = { (y,z) in w x Uw | R(y,z) }
Then R' satisfies the three conditions for AC. However, AC does not imply Global AC. The fact that we have choice functions for each set does not entail, near as I can figger, a choice function for the class of all sets.
>> On Nov 10, 7:53 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote: >>> Charlie-Boo <shymath...@gmail.com> writes: >>> >> Tell you what. Why don't you write down the axiom of choice and point >>> >> out where it involves such an R?
>>> > Let aoc() be the choice function. Then aoc(x)=y iff R(x,y).
>>> Wow. What an utter failure to write down the axiom of choice. Want >>> to try again?
>>> You speak, after all, as if there is a single choice function. Tain't >>> so.
>> I know. That's good of you to understand the set axioms so well. I >> like the simpler version. So, do we know that they aren't >> equivalent?
> As I just posted Global AC (your simpler conditions) imply AC, but > there is no reason to think that AC implies Global AC as far as I > know.
> I'd imagine that the proof that countable choice does not imply AC > gives a hint as to how one would show Global AC does not imply AC,
^^^^^^^^^^^^^^^^^^^^^^^^^^^
> but I'm not familiar with that argument.
Sorry, I meant to say "AC does not imply Global AC".
-- Jesse F. Hughes "My baby don't allow me in the kitchen and I've come to love her decision." -- Bad Livers
> Charlie-Boo <shymath...@gmail.com> writes: > > On Nov 10, 7:51 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote: > >> Charlie-Boo <shymath...@gmail.com> writes: > >> >> Sorry, still not clear on what you mean. The axiom of choice does not > >> >> involve a relation R that you described.
> >> > Ok. # 1 = AOC doesn't involve my R.
> >> >> The relation R that you > >> >> described would be more like what one would see in an axiom of choice > >> >> for classes.
> >> > Ok. # 2 = AOC involves my R when talking about classes.
> >> > But # 1 => ~(# 2).
> >> You're talking nonsense. The axiom of choice refers to a particular > >> (equivalence class of) axiom(s). It is an axiom about sets. The > >> class-based axiom of choice that I mentioned (which I've never seen in > >> the literature) is a different axiom.
> >> Thus #1 does not entail ~(#2).
> > I guess it depends on your definition of "involves". It is a very > > broad word to me.
> They are related, but I wouldn't say that AC involves your R.
> > Anyway, how is R about classes and not sets? You may be getting to > > the point, actually. Classes are for things that are not sets > > (=relations) so AOC is really about whether R is a set. Russell > > proved that some things aren't sets and I am trying to apply > > additional logic to address R being a set or not.
> I shouldn't have said that R was about classes per se,
Ok.
> but your conditions claim that there is essentially a *global* choice function, > that is a choice function for the particular class V.
How is that a but - what does it have to do with R being about classes per se?
> That's not what > AC says.
But that's what I say! That's what CBL says, too. (And CBL proves all sorts of theorems very easily and amazingly short, due to several subterfuges in use.)
Why can't we say that instead?
The question is (as you discuss) whether AC = Global AC. At the least let us add that to the questions discussed, in the mainstream literature (full of fraud) as well as this counter-technology that we are now all collectively developing in a huge collaboration. (Billions access Google.)
But 1st things 1st - I asked you first - is there such an R? (Is it close enough to what books with pretty covers talk about?)
The first problem with ZF addressing AOC is that ZF doesn't define what a function is - there are NO REFERENCES to them - so naturally ZF is consistent with AOC. **
> For all w, there is an R c w x Uw such that the following three > conditions hold:
> (Ay in w)( (Ex)( x in y ) -> (Ez) R(y,z) ) > (Ay)(Az)( R(y,z) -> z in y ) > (Ay)(Az)(Az') ( ( R(y,z) & R(y,z') ) -> z = z' )
> Global AC:
> There is an R such that the following three conditions hold:
> (Ay)( (Ex)( x in y ) -> (Ez) R(y,z) ) > (Ay)(Az)( R(y,z) -> z in y ) > (Ay)(Az)(Az') ( ( R(y,z) & R(y,z') ) -> z = z' )
> Those are two different claims. In Global AC, it is clear that R > cannot be a set at all. It must be a proper class of ordered pairs.
> Clearly, Global AC implies AC. Suppose w is a set and let R be given > as in Global AC. Define
> R' = { (y,z) in w x Uw | R(y,z) }
> Then R' satisfies the three conditions for AC. However, AC does not > imply Global AC. The fact that we have choice functions for each set > does not entail, near as I can figger, a choice function for the class > of all sets.
> -- > Jesse F. Hughes
> Baba: Spell checkers are bad. > Quincy (age 7): C-H-E-K-E-R-S A-R-E B-A-D.- Hide quoted text -
> Charlie-Boo <shymath...@gmail.com> writes: > > On Nov 10, 7:53 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote: > >> Charlie-Boo <shymath...@gmail.com> writes: > >> >> Tell you what. Why don't you write down the axiom of choice and point > >> >> out where it involves such an R?
> >> > Let aoc() be the choice function. Then aoc(x)=y iff R(x,y).
> >> Wow. What an utter failure to write down the axiom of choice. Want > >> to try again?
> >> You speak, after all, as if there is a single choice function. Tain't > >> so.
> > I know. That's good of you to understand the set axioms so well. I > > like the simpler version. So, do we know that they aren't > > equivalent?
> As I just posted Global AC (your simpler conditions) imply AC, but > there is no reason to think that AC implies Global AC as far as I > know.
> I'd imagine that the proof that countable choice does not imply AC
I'll savor this one and give you a chance. (Also upping the ante.) But also maybe we're onto something big (relatively.) Someone proved that something doesn't imply AC? And what could we conclude from that little morsel? (I realized this only on my second reading.)
> gives a hint as to how one would show Global AC does not imply AC, but > I'm not familiar with that argument.
> -- > Jesse F. Hughes > "Yes, I'm one of those arrogant people who tries to be quotable. > There is actually at least one person who quotes me often." > -- James Harris- Hide quoted text -
> Charlie-Boo <shymath...@gmail.com> writes: > > On Nov 10, 7:51 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote: > >> Charlie-Boo <shymath...@gmail.com> writes: > >> >> Sorry, still not clear on what you mean. The axiom of choice does not > >> >> involve a relation R that you described.
> >> > Ok. # 1 = AOC doesn't involve my R.
> >> >> The relation R that you > >> >> described would be more like what one would see in an axiom of choice > >> >> for classes.
> >> > Ok. # 2 = AOC involves my R when talking about classes.
> >> > But # 1 => ~(# 2).
> >> You're talking nonsense. The axiom of choice refers to a particular > >> (equivalence class of) axiom(s). It is an axiom about sets. The > >> class-based axiom of choice that I mentioned (which I've never seen in > >> the literature) is a different axiom.
> >> Thus #1 does not entail ~(#2).
> > I guess it depends on your definition of "involves". It is a very > > broad word to me.
> They are related, but I wouldn't say that AC involves your R.
A is related to B but A does not involve B?
> > Anyway, how is R about classes and not sets? You may be getting to > > the point, actually. Classes are for things that are not sets > > (=relations) so AOC is really about whether R is a set. Russell > > proved that some things aren't sets and I am trying to apply > > additional logic to address R being a set or not.
> I shouldn't have said that R was about classes per se,
I don't even think you should say AC and R are related but noninvolving.
> but your > conditions claim that there is essentially a *global* choice function, > that is a choice function for the particular class V.
I also don't think you should say but.
However, I do think you should think about using Theory of Computation proofs of completeness and incompleteness to prove R exists or not, to address AOC. (As long as I get 1/2 of the prize money. (What's it up to?))
The first question could be (start with the simple stuff - substitution), what about R if we substitute YES for SE in the definition of R? Anybody?
Plz excuse me for a few hours or days while I switch gears from being the first to prove (orchestrate) that AOC is impossible, to being the first to write an algorithm (as evidenced by its nonexistance on the internet) for the world's first HTML to SQL translator (speaking of formalizing and automating things.)
> For all w, there is an R c w x Uw such that the following three > conditions hold:
> (Ay in w)( (Ex)( x in y ) -> (Ez) R(y,z) ) > (Ay)(Az)( R(y,z) -> z in y ) > (Ay)(Az)(Az') ( ( R(y,z) & R(y,z') ) -> z = z' )
> Global AC:
> There is an R such that the following three conditions hold:
> (Ay)( (Ex)( x in y ) -> (Ez) R(y,z) ) > (Ay)(Az)( R(y,z) -> z in y ) > (Ay)(Az)(Az') ( ( R(y,z) & R(y,z') ) -> z = z' )
> Those are two different claims. In Global AC, it is clear that R > cannot be a set at all. It must be a proper class of ordered pairs.
> Clearly, Global AC implies AC. Suppose w is a set and let R be given > as in Global AC. Define
> R' = { (y,z) in w x Uw | R(y,z) }
> Then R' satisfies the three conditions for AC. However, AC does not > imply Global AC. The fact that we have choice functions for each set > does not entail, near as I can figger, a choice function for the class > of all sets.
> -- > Jesse F. Hughes
> Baba: Spell checkers are bad. > Quincy (age 7): C-H-E-K-E-R-S A-R-E B-A-D.- Hide quoted text -
> However, AC does not imply Global AC. The fact that we have choice > functions for each set does not entail, near as I can figger, a choice > function for the class of all sets.
Your figgering can be backed up with a logical result. It is also a logical result, an easy and illustrative application of forcing, that any invocation of global choice in a proof of a result about sets only can be eliminated (given ordinary choice).
Charlie-Boo <shymath...@gmail.com> writes: > On Nov 10, 9:34 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote: >> Charlie-Boo <shymath...@gmail.com> writes: >> > On Nov 10, 7:51 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote: >> >> Charlie-Boo <shymath...@gmail.com> writes: >> >> >> Sorry, still not clear on what you mean. The axiom of choice does not >> >> >> involve a relation R that you described.
>> >> > Ok. # 1 = AOC doesn't involve my R.
>> >> >> The relation R that you >> >> >> described would be more like what one would see in an axiom of choice >> >> >> for classes.
>> >> > Ok. # 2 = AOC involves my R when talking about classes.
>> >> > But # 1 => ~(# 2).
>> >> You're talking nonsense. The axiom of choice refers to a particular >> >> (equivalence class of) axiom(s). It is an axiom about sets. The >> >> class-based axiom of choice that I mentioned (which I've never seen in >> >> the literature) is a different axiom.
>> >> Thus #1 does not entail ~(#2).
>> > I guess it depends on your definition of "involves". It is a very >> > broad word to me.
>> They are related, but I wouldn't say that AC involves your R.
> A is related to B but A does not involve B?
Let us not quibble on such dull matters of terminology.
>> > Anyway, how is R about classes and not sets? You may be getting to >> > the point, actually. Classes are for things that are not sets >> > (=relations) so AOC is really about whether R is a set. Russell >> > proved that some things aren't sets and I am trying to apply >> > additional logic to address R being a set or not.
>> I shouldn't have said that R was about classes per se,
> I don't even think you should say AC and R are related but > noninvolving.
>> but your >> conditions claim that there is essentially a *global* choice function, >> that is a choice function for the particular class V.
> I also don't think you should say but.
> However, I do think you should think about using Theory of Computation > proofs of completeness and incompleteness to prove R exists or not, to > address AOC. (As long as I get 1/2 of the prize money. (What's it up > to?))
> The first question could be (start with the simple stuff - > substitution), what about R if we substitute YES for SE in the > definition of R? Anybody?
> Plz excuse me for a few hours or days while I switch gears from being > the first to prove (orchestrate) that AOC is impossible, to being the > first to write an algorithm (as evidenced by its nonexistance on the > internet) for the world's first HTML to SQL translator (speaking of > formalizing and automating things.)
No idea what you're going on about.
-- Jesse F. Hughes
"I am the next legend--living, breathing and solving mega problems in the here and now." -- James S. Harris
Aatu Koskensilta <aatu.koskensi...@uta.fi> writes: > "Jesse F. Hughes" <je...@phiwumbda.org> writes:
>> However, AC does not imply Global AC. The fact that we have choice >> functions for each set does not entail, near as I can figger, a choice >> function for the class of all sets.
> Your figgering can be backed up with a logical result. It is also a > logical result, an easy and illustrative application of forcing, that > any invocation of global choice in a proof of a result about sets only > can be eliminated (given ordinary choice).
It is, of course, only coincidental when my figgering and the truth line up so well.
-- "Tempted and tried we're oft made to wonder Why it should be thus all the day long When there are others living about us Never molested though in the wrong." -- Bad Livers, "Farther Along"
Charlie-Boo <shymath...@gmail.com> writes: > On Nov 10, 9:34 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote: >> Charlie-Boo <shymath...@gmail.com> writes: >> > Anyway, how is R about classes and not sets? You may be getting to >> > the point, actually. Classes are for things that are not sets >> > (=relations) so AOC is really about whether R is a set. Russell >> > proved that some things aren't sets and I am trying to apply >> > additional logic to address R being a set or not.
>> I shouldn't have said that R was about classes per se,
> Ok.
>> but your conditions claim that there is essentially a *global* choice function, >> that is a choice function for the particular class V.
> How is that a but - what does it have to do with R being about classes > per se?
Your conditions of R amount to the axiom of choice for a particular class, not for all classes of sets.
>> That's not what >> AC says.
> But that's what I say! That's what CBL says, too. (And CBL proves > all sorts of theorems very easily and amazingly short, due to several > subterfuges in use.)
But who cares what you say? You said that lots of people are writing about relations R satisfying those three conditions. That's just not true.
> Why can't we say that instead?
> The question is (as you discuss) whether AC = Global AC. At the least > let us add that to the questions discussed, in the mainstream > literature (full of fraud) as well as this counter-technology that we > are now all collectively developing in a huge collaboration. > (Billions access Google.)
Well, you can certainly ask that question. Seems to me that the answer is almost certainly "no", but I haven't a proof of that fact. Do you have any argument why the answer may be "yes"?
> But 1st things 1st - I asked you first - is there such an R? (Is it > close enough to what books with pretty covers talk about?)
In the theory ZFC, certainly not (because R would not be a set). If we amend ZFC so that it makes sense to speak of proper classes, then I have no proof that there is no class R satisfying your conditions. Nor do I have a proof that there is such a class. Moreover, I sincerely doubt that the latter claim is provable (though I haven't an argument to that effect).
> The first problem with ZF addressing AOC is that ZF doesn't define > what a function is - there are NO REFERENCES to them - so naturally ZF > is consistent with AOC. **
I'm not sure what you're going on about. The axioms of ZF do not define function, but it is easy enough to introduce such a definition. Here it is:
Let f, X and Y be sets. Then f is a function with domain X and codomain Y (written f:X -> Y) iff the following hold:
(1) f c X x Y (f is a subset of X x Y) (2) (Ax in X)(Ey in Y)( <x,y> in f ) (3) (Ax in X)(Ay in Y)(Ay' in Y)( ( <x,y> in f & <x,y'> in f ) -> y = y' )
>> For all w, there is an R c w x Uw such that the following three >> conditions hold:
>> (Ay in w)( (Ex)( x in y ) -> (Ez) R(y,z) ) >> (Ay)(Az)( R(y,z) -> z in y ) >> (Ay)(Az)(Az') ( ( R(y,z) & R(y,z') ) -> z = z' )
>> Global AC:
>> There is an R such that the following three conditions hold:
>> (Ay)( (Ex)( x in y ) -> (Ez) R(y,z) ) >> (Ay)(Az)( R(y,z) -> z in y ) >> (Ay)(Az)(Az') ( ( R(y,z) & R(y,z') ) -> z = z' )
>> Those are two different claims. In Global AC, it is clear that R >> cannot be a set at all. It must be a proper class of ordered pairs.
>> Clearly, Global AC implies AC. Suppose w is a set and let R be given >> as in Global AC. Define
>> R' = { (y,z) in w x Uw | R(y,z) }
>> Then R' satisfies the three conditions for AC. However, AC does not >> imply Global AC. The fact that we have choice functions for each set >> does not entail, near as I can figger, a choice function for the class >> of all sets.
-- "It's my belief that when religion and pseudoscience achieve an official status within a culture [...], then genocide, war, oppression, injustice, and economic stagnation are sure to follow." -- David Petry, on why |X| < |P(X)| is bad, bad, bad.
Charlie-Boo <shymath...@gmail.com> writes: > I'll savor this one and give you a chance. (Also upping the ante.) > But also maybe we're onto something big (relatively.) Someone proved > that something doesn't imply AC? And what could we conclude from that > little morsel? (I realized this only on my second reading.)
The axiom of choice is independent of ZF, you know. Thus, someone proved that ZF does not imply AC (also, that ZF does not imply ~AC).
It has also been proved that ZF + CC is independent of AC. Thus, ZF + CC does not prove AC (nor its negation).
No idea what you're going to conclude from this little morsel. I conclude a few things (ZF + ~AC is equiconsistent to ZF, and so is ZF + AC, for instance), but all of my conclusions are obvious and well-known.
-- "There are people [...] who think it's socially acceptable to level accusations of mental illness in insulting exchanges to make points[...] [They] are rather sick [them]selves, and in reality, are sociopathic." --- James Harris, evidently a self-described sociopath
In message <87ocnap6x6....@phiwumbda.org>, Jesse F. Hughes <je...@phiwumbda.org> writes
>In the theory ZFC, certainly not (because R would not be a set). If we >amend ZFC so that it makes sense to speak of proper classes, then I >have no proof that there is no class R satisfying your conditions. Nor >do I have a proof that there is such a class. Moreover, I sincerely >doubt that the latter claim is provable (though I haven't an argument >to that effect).
"V=L" provides a universal well-ordering and so a universal choice function. So it is consistent with ZFC (+ classes) that such an R exists. -- David Hartley
David Hartley <m...@privacy.net> writes: > In message <87ocnap6x6....@phiwumbda.org>, Jesse F. Hughes > <je...@phiwumbda.org> writes >>In the theory ZFC, certainly not (because R would not be a set). If we >>amend ZFC so that it makes sense to speak of proper classes, then I >>have no proof that there is no class R satisfying your conditions. Nor >>do I have a proof that there is such a class. Moreover, I sincerely >>doubt that the latter claim is provable (though I haven't an argument >>to that effect).
> "V=L" provides a universal well-ordering and so a universal choice > function. So it is consistent with ZFC (+ classes) that such an R > exists.
Thanks for the clarification. Thus, there is no proof that the class R does not exist (in ZFC + classes). And, if I understood Aatu's post, there is similarly no proof that the class R exists. Hence, Global AC is independent of ZFC + classes.
I hope I got that right.
If so, surely, that is the answer to Charlie's question (though not, I'd wager, the answer he wanted to receive).
-- Jesse F. Hughes
"The sole cause of all human misery is the inability of people to sit quietly in their rooms." -- Blaise Pascal
Jesse F. Hughes wrote: > In the theory ZFC, certainly not (because R would not be a set). If > we amend ZFC so that it makes sense to speak of proper classes, ...
Is there a difference between 'ZFC with classes' and NBG?
