Uncountable sets in CZF?
Agamemnon
mathematics and I was hoping someone here could answer a set theory
question that I have. When I first heard about uncountable sets, I
thought it was very strange that some infinite sets are "larger" than
others. It troubled me at first, but eventually I thought to myself
"All this means is that if you define size (cardinality) a certain
way, and use these particular axioms to capture the logic of sets,
then it follows that uncountable sets exist. Don't read too much into
it."
It seems to me that in a constructive set theory, such as CZF, it
would not be possible to prove the existence of a set with a
cardinality greater than that of the natural numbers. I say this
because, as far as I know, proofs of cantor's theorem (and similar
theorems) in ZFC are always non-constructive. Does anyone know if this
is true?
Note: I'm not claiming that CZF is superior to ZFC or that ZFC is in
some way flawed, so please don't flame me for that.
William Elliot
> It seems to me that in a constructive set theory, such as CZF, it
> would not be possible to prove the existence of a set with a
> cardinality greater than that of the natural numbers. I say this
> because, as far as I know, proofs of cantor's theorem (and similar
> theorems) in ZFC are always non-constructive. Does anyone know if this
> is true?
>
Indeed you can construct the reals, but can you constructive show
|N| < |R| ?
It's easy to construct a proof of
|N| <= |R|
yet more is needed, namely
|N| /= |R|
That's quite a conundrum in itself, to constructively prove a negation.
I would think not. That
|N| /= |R|
is not a theorem of CZF.
You may enjoy the Loewenheim-Skolem paradox, that any countable FOL
sufficient for constructing uncountable sets, has a countable model.
fishfry
agamemno...@yahoo.com (Agamemnon) wrote:
> Hello everyone. I am a philosophy student with an interest in
> mathematics and I was hoping someone here could answer a set theory
> question that I have. When I first heard about uncountable sets, I
> thought it was very strange that some infinite sets are "larger" than
> others. It troubled me at first, but eventually I thought to myself
> "All this means is that if you define size (cardinality) a certain
> way, and use these particular axioms to capture the logic of sets,
> then it follows that uncountable sets exist. Don't read too much into
> it."
>
Yes, but these axioms and definitions have a certain intuitive appeal.
You agree that some infinite sets can be put into one-one correspondence
with the natural numbers; and other infinite sets, can't. You must still
regard that as strange and wondrous, no? Because the definitions and
axioms are in no way artificial or forced.
Torkel Franzen
> That's quite a conundrum in itself, to constructively prove a
> negation.
A constructive proof of a negative statement -A consists in a
derivation of a contradiction from A.
Aatu Koskensilta
> I say this because, as far as I know, proofs of cantor's theorem (and similar
> theorems) in ZFC are always non-constructive. Does anyone know if this
> is true?
It's not. The usual proof of Cantor's theorem does not rely on
non-constructive modes of inference. As Torkel explained, a proof of
contradiction from A is a constructive proof of ~A.
--
Aatu Koskensilta (aatu.kos...@xortec.fi)
"Wovon man nicht sprechen kann, daruber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
Jesse F. Hughes
> Hello everyone. I am a philosophy student with an interest in
> mathematics and I was hoping someone here could answer a set theory
> question that I have. When I first heard about uncountable sets, I
> thought it was very strange that some infinite sets are "larger" than
> others. It troubled me at first, but eventually I thought to myself
> "All this means is that if you define size (cardinality) a certain
> way, and use these particular axioms to capture the logic of sets,
> then it follows that uncountable sets exist. Don't read too much into
> it."
>
> It seems to me that in a constructive set theory, such as CZF, it
> would not be possible to prove the existence of a set with a
> cardinality greater than that of the natural numbers. I say this
> because, as far as I know, proofs of cantor's theorem (and similar
> theorems) in ZFC are always non-constructive. Does anyone know if this
> is true?
No, it's not true.
Theorem: For any set X, there is no surjection X -> PX.
Proof:
Let X and f:X -> PX be given and we want to show that f is not a
surjection. As usual, define
R = { x in X | x is not in f(x) }
and we will show that it is *not* the case that there is an x such
that f(x) = R.
As Torkel says, to derive a negation constructively, one shows that
the unnegated formula yields a contradiction. Negation introduction
is a constructive rule of inference, but negation elimination isn't.
Thus, suppose that there is an x such that f(x) = R. Suppose that x
is in f(x). Then x is not in f(x), which is a contradiction and
thus we conclude NOT( x in f(x) ).
We've proved, then, that x is not in f(x) and so it must be that x
doesn't satisfy the defining predicate for R --- that is,
NOT( NOT ( x in f(x) ) )
Hence, NOT( x in f(x) ) and NOT( NOT ( x in f(x) ) ). We have thus
proved a contradiction and my conclude that there is no x such that
f(x) = R.
See? You don't need excluded middle or negation elimination for this
proof at all. If you want to be certain I didn't do anything funky,
just formalize this outline in your favorite intuitionistic logic (I
assume that for our purposes, intuitionism is more or less the same as
constructivism --- but just for the purposes of your question).
--
Jesse F. Hughes
"What I represent is the unknowable future--the power of change. In
that sense I'm a force of Nature, a force of the Universe, a living
emodiment of change itself." --James Harris and his sense of humility
William Elliot
Oh, so a constructive proof of A, cannot be to show ~A gives a
contradiction. All that does is show ~~A. Hm...
Herman Jurjus
> agamemno...@yahoo.com (Agamemnon) writes:
>
>
>>Hello everyone. I am a philosophy student with an interest in
>>mathematics and I was hoping someone here could answer a set theory
>>question that I have. When I first heard about uncountable sets, I
>>thought it was very strange that some infinite sets are "larger" than
>>others. It troubled me at first, but eventually I thought to myself
>>"All this means is that if you define size (cardinality) a certain
>>way, and use these particular axioms to capture the logic of sets,
>>then it follows that uncountable sets exist. Don't read too much into
>>it."
>>
>>It seems to me that in a constructive set theory, such as CZF, it
>>would not be possible to prove the existence of a set with a
>>cardinality greater than that of the natural numbers. I say this
>>because, as far as I know, proofs of cantor's theorem (and similar
>>theorems) in ZFC are always non-constructive. Does anyone know if this
>>is true?
>
>
> No, it's not true.
>
> Theorem: For any set X, there is no surjection X -> PX.
[snip]
> See? You don't need excluded middle or negation elimination for this
> proof at all. If you want to be certain I didn't do anything funky,
> just formalize this outline in your favorite intuitionistic logic (I
> assume that for our purposes, intuitionism is more or less the same as
> constructivism --- but just for the purposes of your question).
Did you ever meet an intuitionist who accepts the existence of 'the
power set' of an infinite set? (I know intuitionists who even refuse to
use the word 'set'.)
--
Cheers,
Herman Jurjus
Jesse F. Hughes
Well, when I used the word above, all I meant was set theory in which
the underlying logic is intuitionistic. I was speaking about
mathematical logic, not philosophical foundations.
--
Jesse F. Hughes
"What does soap kill? Germs or Germans?"
-- Quincy P. Hughes (age 3 1/2) asks for clarification
Ross A. Finlayson
Hi fishfry. Strange and wondrous? This is mathematics. Cut it open
and look inside. There are rare cases where that makes a difference.
There are no paradoxes. Is thus everything a paradox? I disagree
about the axioms.
I've read recently, as have probably you, that in a model of an
intuitionist zet theory, IZF for intuitionist Zermelo-Fraenkel, that
it is not inconsistent for there to be a mapping between set and
powerset. That agrees with some of the things I say.
We had a good discussion within a month about that, see the thread
"'Uncountable' doesn't exist." That contains some further discussion
of this issue by various authors that send copies of their words to
sci.math, among them highly skilled amateur and professional
mathematicians.
Imagine if a set theory had a set of all sets. Then, that set would
be its own powerset.
We've been discussing models of ubiquitous ordinals. In one of those
models EF is bijection between N and the each element of the unit
interval.
I'd like to know more about what Loewenheim-Skolem is saying. Maybe
it can explain to Banach-Tarski that a point on a line has two sides,
but only one on either side. Plain language?
Menzel's description of the halting problem (The Halting Problem, THP)
on sci.logic was worth reading.
Unitize the analog!
Ross
Andrew Usher
> Hello everyone. I am a philosophy student with an interest in
> mathematics and I was hoping someone here could answer a set theory
> question that I have. When I first heard about uncountable sets, I
> thought it was very strange that some infinite sets are "larger" than
> others. It troubled me at first, but eventually I thought to myself
> "All this means is that if you define size (cardinality) a certain
> way, and use these particular axioms to capture the logic of sets,
> then it follows that uncountable sets exist. Don't read too much into
> it."
You might want to read some of the thread "'Uncountable' doesn't
exist', from a few weeks ago. What we know is that some classes of
things are larger than any infinite enumeration, such as the real
numbers. Calling them a 'set' is a large leap of faith, but if you do,
then yes, they are uncountable.
> It seems to me that in a constructive set theory, such as CZF, it
> would not be possible to prove the existence of a set with a
> cardinality greater than that of the natural numbers. I say this
> because, as far as I know, proofs of cantor's theorem (and similar
> theorems) in ZFC are always non-constructive. Does anyone know if this
> is true?
What additional axiom does CZF have? If it is consistent, it would
disprove the claim of Jesse Hughes that V>=L necessarily.
> Note: I'm not claiming that CZF is superior to ZFC or that ZFC is in
> some way flawed, so please don't flame me for that.
Both are _consistent_. If you're a Platonist, only one theory can be
actually _true_, however.
Andrew Usher
Jesse F. Hughes
> I've read recently, as have probably you, that in a model of an
> intuitionist zet theory, IZF for intuitionist Zermelo-Fraenkel, that
> it is not inconsistent for there to be a mapping between set and
> powerset. That agrees with some of the things I say.
This is false, if by "mapping" you mean "bijection".
You should choose what you read more carefully. In this thread, I
gave an explicit proof that in IZF, there is no surjection from X to
PX for any set X.
--
"And yes, for those who think that just maybe I did find a short proof
of Fermat's Last Theorem, and THE prime counting function, if I
succeed at what I'm working on now world economy as you know it will
be gone." -- James Harris branches out.
Jesse F. Hughes
> What additional axiom does CZF have? If it is consistent, it would
> disprove the claim of Jesse Hughes that V>=L necessarily.
You are badly confused.
I know that I didn't respond to your question in
<6e197594.04081...@posting.google.com>, but I was hoping
someone that knows the definition of non-measurable would give it a
go. I'll give it my best nonetheless.
You wrote:
,----[ <6e197594.04081...@posting.google.com> ]
| Does the construction of L require AC to begin with?
|
| To go back to a previous example, AC -> there is a non-measurable
| subset of R.
|
| If V=L -> AC, then L must contain such a set.
|
| Clearly, adding more sets to V (and making AC false) can't eliminate
| anything, so there must be a non-measurable regardless of AC, if L is
| contained in V, right?
`----
I don't know diddly about the definitions, but I imagine that the
definition of non-measurable includes some quantifiers. If we enlarge
the universe of discourse, then a set which was previously
non-measurable may well become measurable. Someone that knows a
definition or two can confirm or deny this, but I'd reckon that a set
is not measurable/non-measurable all by its lonesome, but only in some
context, namely the universe.
In any case, it is just painfully obvious that V >= L. L is the
collection of all sets which provably exist. Any model of ZF must
contain every set which provably exists (well, perhaps with some of
these sets identified). Let's call a model M of ZF /good/ if,
whenever M |= s = t for any terms s and t, then also ZF |- s = t.
Some model theorist can tell me what the right terminology is, but
"good" will do for now. Clearly, if M is good, then L <= M. That is,
then L is a submodel of M.
V is clearly a good model of ZF. Therefore L is a submodel (not
necessarily proper) of V.
This is more detail than ought to be necessary. Every constructible
set is a *set*. Therefore, L is contained in V. Duh.
--
"But he himself was not to blame for his vices. They grew out of a personal
defect in his mother. She did her best in the way of flogging him while an
infant... but, poor woman! she had the misfortune to be left-handed, and a
child flogged left-handedly had better be left unflogged." -- E.A. Poe
Agamemnon
Then I had a misunderstanding about intuitionistic logic. I figured
that without the law of excluded middle, it would be possible for both
A and -A to be false. Deriving a contradiction from A would prove A
false, but that wouldn't be the same as proving -A true (or so I
thought).
However, I just looked at a list of axioms for intuitionistic logic,
and it included the following: (A -> B) -> ((A -> -B) -> -A). So it
seems you can prove -A by deriving a contradiction from A, but you
can't prove A by deriving a contradiction from -A (you'd just prove
--A). That's what confused me.
Since Cantor's theorem is a negative, I guess you can prove it even in
CZF. This still does not answer the question of whether or not
uncountable sets exist in CZF. After all, CZF does not have the power
set axiom, but it must have something similar since CZF is clasicaly
equivalent to ZF.
So... back to my original question: do constructive set theories in
general, or CZF in particular, have uncountable sets (i.e. can they be
proven to exist)?
Thanks for your time everyone.
-Agamemnon
Agamemnon
[snip]
>
> See? You don't need excluded middle or negation elimination for this
> proof at all. If you want to be certain I didn't do anything funky,
> just formalize this outline in your favorite intuitionistic logic (I
> assume that for our purposes, intuitionism is more or less the same as
> constructivism --- but just for the purposes of your question).
Thanks for the post. Actually I was already familar with this proof, I
just misunderstood what forms of proof are considered constructive. I
knew that you couldn't prove A from deriving a contradiction from -A,
but I didn't realize you CAN prove -A by deriving a contradiction from
A. Thanks for clearing that up.
-Agamemnon
Torkel Franzen
> So... back to my original question: do constructive set theories in
> general, or CZF in particular, have uncountable sets (i.e. can they be
> proven to exist)?
As for CZF, apparently not. But I'm unfamiliar with the subject, and
you can check it out for yourself:
www.ml.kva.se/preprints/ archive/2000-2001/2000-2001-40.ps
(Notes on Constructive Set Theory by Aczel and Rathjen.)
Herman Jurjus
Just FYI: intuitionists like to think of their stuff as mathematics
('intuitionitic mathematics'), not as logic or as philosophy.
Coming back to the original question: in (some forms of) intuitionist
mathematics, N has uncountable subsets, so 'larger size' has very little
to do with uncountability. It is rather odd that the arguments typically
given for Cantor's theorem are constructively acceptable, but the
conclusion is not even understandable, intuitionistically (at least in
the more interesting variants). Even if you want to ignore such kinds of
intuitionist mathematics as weird or irrelevant for 'real mathematics',
doesn't this suggest that the classical argument sneaks in a number of
hidden assumptions somewhere, and keeps unacceptably silent about it?
Which may explain why so many people initially feel that there is
something wrong with it. And this may also explain why the op thinks all
proofs are 'non-constructive'. Just speculating...
--
Cheers,
Herman Jurjus
Agamemnon
> > "All this means is that if you define size (cardinality) a certain
> > way, and use these particular axioms to capture the logic of sets,
> > then it follows that uncountable sets exist. Don't read too much into
> > it."
> >
>
> Yes, but these axioms and definitions have a certain intuitive appeal.
> You agree that some infinite sets can be put into one-one correspondence
> with the natural numbers; and other infinite sets, can't. You must still
> regard that as strange and wondrous, no? Because the definitions and
> axioms are in no way artificial or forced.
It is interesting, but I wouldn't use the word 'wondrous'. I agree
that, in the real world, two collections have the same size if their
elements can be put into a one-to-one correspondence. However I don't
agree that my intuitive notion of a one-to-one correspondence which I
use in the real world is the same as a set theorist's notion of a
bijection in set theory.
In the real world, I can show a one-to-one correspondence of apples
and oranges by placing each apple next to exactly one orange. Within
set theory, I cannot place sets next to each other, or draw an arrow
from one set to another - the only way to show the correspondence is
by using other sets within the theory.
Given finitly many symbols, one can make only countably many
sentences, which means one can only define countably many real
numbers. But from within set theory you can prove that there are
uncountably many reals. So do we conclude that some reals are
undefinable? Remember, it is impossible to come up with a
counter-example to the claim "all reals are defineable." Or do we
conclude that our notion of a bijection within set theory isn't the
same as our intuitive notion of a one-to-one correspondence in the
real world, and so we cannot meaningfully speak about the relative
sizes of infinite collections?
Oh, and as for the axioms of ZFC not being 'artificial', I'm not so
sure about that. In a natural language such as English, people can
talk about collections without restriction, and this leads to various
paradoxes. I think our intuitive notion of a set is contradictory -
just like Cantor's "naive" set theory. In order to axiomatize the
concept of sets in a non-contradictory way, restrictions must be
placed on how we can make sets. But ZFC is only one way of doing this
and so it is, in that sense, artificial. Sure, the axioms of ZFC are
pretty intuitive, but so are the axioms of naive set theory.
Quine's NF trys to be more like naive set theory, and many theorems of
ZFC are false in NF (Cantor's theorem is one example). CZF, on the
other hand, is even more restrictive than ZFC. I suspect many of the
"paradoxes" of ZFC do not occur in CZF, but I'm not sure about this.
(the Banach-Tarski paradox is one example of the sort of paradox I'm
talking about)
-Agamemnon
Torkel Franzen
> I think our intuitive notion of a set is contradictory -
> just like Cantor's "naive" set theory.
Cantor's theory was not at all contradictory. On the contrary, all
of his mathematics stands today.
> In order to axiomatize the
> concept of sets in a non-contradictory way, restrictions must be
> placed on how we can make sets.
Or we need to explain more clearly what we are talking about, as in
the iterative conception of set.
Jesse F. Hughes
> Just FYI: intuitionists like to think of their stuff as mathematics
> ('intuitionitic mathematics'), not as logic or as philosophy.
There is intuitionism as a branch of mathematical logic and
intuitionism as a mathematical foundation.
If one is only interested in intuitionism as one in a family of
alternative logics, then he wouldn't balk at the powerset axiom. It's
just another axiom. Such people aren't usually called
"intuitionists" in my experience.
If one advocates that intuitionism is the *right* way to do
mathematics, then he's advocating a philosophical position. In this
case, he is very likely to object to taking powersets of infinite
sets.
At least, that is how I apply the terms.
--
"Now I realize that he got away with all of that because sci.math is
not important, and the rest of the world doesn't pay attention.
Like, no one is worried about football players reading sci.math
postings!" -- James S. Harris on jock reading habits
Jesse F. Hughes
>> In order to axiomatize the
>> concept of sets in a non-contradictory way, restrictions must be
>> placed on how we can make sets.
>
> Or we need to explain more clearly what we are talking about, as in
> the iterative conception of set.
Do you think that the iterative conception of set is a clarification
of the previous naive set theory, or a real restriction on how sets
are constructed?
I'd guess the latter, but I'm not really sure. I just don't see how
the notion of set so strongly suggests the iterative concept that its
omission in naive set theory was essentially an oversight. And I
think that the success of anti-well-founded set theory casts further
doubt that the iterative conception is just a matter of "explain[ing]
more clearly what we [were] talking about."
--
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
Thanks for the reference, Torkel. Finally, I can see what the heck
people are on about around here. Very useful.
--
"These mathematicians are worse than communists, as how do you explain
their behavior? I *am* the American Dream, fighting for what should be
mine, having to get past weak-minded academics who are fighting to
block my success. But I shall prevail!!!" -- James S. Harris
Torkel Franzen
> Do you think that the iterative conception of set is a clarification
> of the previous naive set theory, or a real restriction on how sets
> are constructed?
That's a historical question, and I don't know the answer. As far as
I know, it was only Zermelo who (in 1930) stated clearly the iterative
conception.
Agamemnon
>
> Cantor's theory was not at all contradictory. On the contrary, all
> of his mathematics stands today.
Are you unfamiliar with Russell's paradox? Or are you claiming that
Cantor's set theory did not actually suffer from it?
Torkel Franzen
> Are you unfamiliar with Russell's paradox? Or are you claiming that
> Cantor's set theory did not actually suffer from it?
Cantor's set theory did not suffer one little bit from Russell's
paradox.
Chairman of the Ozzy Osbourne Appreciation Society
Is the comprehension principle part of Cantor's theory?
or is it something which belongs to naive set theory in general,
which isn't necessarily part of Cantor's theory?
--
Replace Roman numerals with digits to reply by email
Torkel Franzen
> Is the comprehension principle part of Cantor's theory?
No.
> or is it something which belongs to naive set theory in general,
No, but it was part of Frege's theory.
Lee Rudolph
>Torkel Franzen <tor...@sm.luth.se> wrote in message news:<vcbzn4s...@beta19.sm.ltu.se>...
>>
>> Cantor's theory was not at all contradictory. On the contrary, all
>> of his mathematics stands today.
>
>Are you unfamiliar with Russell's paradox?
Is Hippodemia unfamiliar with sucking eggs?
Lee Rudolph
Agamemnon
As I understand it, Cantor never gave a rigorous axiomatization of his
set theory, but he implicitly assumed that for any property, there
exists a set of all the objects that have that property. This
unrestricted principle of set comprehension leads directly to
Russell's paradox.
Ross A. Finlayson
super-Cantorian set theory
non-Cantorian set theory
post-Cantorian set theory
Non-Euclidean geometry is also called extra- or super-Euclidean, or Minkowskian.
Ross
Jesse F. Hughes
Well, what theory do you have in mind and I'll tell you what I call
it.
Write down some axioms. Non-Euclidean geometry didn't come about by
the handwaving speculation and diatribes you've offered. It came
about when folks negated the fifth postulate in one of two explicit
ways. Now which part of ZF do you intend to change and how?
--
"My proof has been checked very thoroughly, both by me and others.
Those others apparently decided that they would not believe the proof
was correct, but cannot support that position using mathematics. But
hey, they're just human beings." --JSH, prover of Fermat's Last Thm
Torkel Franzen
> As I understand it, Cantor never gave a rigorous axiomatization of his
> set theory, but he implicitly assumed that for any property, there
> exists a set of all the objects that have that property.
What do you mean by "implicitly assumed"? Since nothing in Cantor's
actual work was invalidated by the paradoxes, you must have in mind
some speculation about mathematically irrelevant private thoughts of
Cantor's. Even such speculations are mistaken, as is clear from
Cantor's letter to Dedekind.
Herman Jurjus
> Herman Jurjus <h.ju...@hetnet.nl> writes:
>
>
>>Just FYI: intuitionists like to think of their stuff as mathematics
>>('intuitionitic mathematics'), not as logic or as philosophy.
>
>
> There is intuitionism as a branch of mathematical logic and
> intuitionism as a mathematical foundation.
>
> If one is only interested in intuitionism as one in a family of
> alternative logics, then he wouldn't balk at the powerset axiom. It's
> just another axiom. Such people aren't usually called
> "intuitionists" in my experience.
>
> If one advocates that intuitionism is the *right* way to do
> mathematics, then he's advocating a philosophical position. In this
> case, he is very likely to object to taking powersets of infinite
> sets.
>
> At least, that is how I apply the terms.
Fair enough.
Just FYI, there is a third possibility, namely to be interested in
intuitionism as one in a family of alternative forms of mathematics.
This can go a lot further than adopting a different set of logical
rules, especially when it comes to set-like theorems.
(There was a time when people were not aware of AC: they silently used
it, and it felt like an inference principle, not as an axiom.)
BTW, studying the mathematical consequences of alternative sets of
pre-assumptions is not the same as defending them as a new 'one and only
right way'. (Quite on the contrary, i would say.)
--
Cheers,
Herman Jurjus
ZZBunker
That's only because as Cantor himself knew,
he didn't HAVE a theory. He had *one* diagonal proof,
that a non-logician called Hilbert later mistakening called
Set heaven. Given that Hilbert knew infinitely
less about power sets than Russell did.
Which is how Russell's Paradox crashed set "theory"
before it even began. Since he was one
of only a very few people, including Goedel, that understood
that Peano Axioms have the peculiar set property
that they are not axioms.
Torkel Franzen
> That's only because as Cantor himself knew,
> he didn't HAVE a theory. He had *one* diagonal proof,
Proof? Do you call Cantor's piece of fraudulent twaddle a proof?
It's sad to see how even progressive thinkers are hoodwinked by
Cantorian propaganda.
Ross A. Finlayson
> r...@tiki-lounge.com (Ross A. Finlayson) writes:
>
> > Which do you prefer:
> >
> > super-Cantorian set theory
> > non-Cantorian set theory
> > post-Cantorian set theory
> >
> > Non-Euclidean geometry is also called extra- or super-Euclidean, or
> > Minkowskian.
>
> Well, what theory do you have in mind and I'll tell you what I call
> it.
>
> Write down some axioms. Non-Euclidean geometry didn't come about by
> the handwaving speculation and diatribes you've offered. It came
> about when folks negated the fifth postulate in one of two explicit
> ways. Now which part of ZF do you intend to change and how?
Hi,
That's a good question.
I think the axioms of ZF minus powerset and regularity are theorems of
a set of zero axioms, where they are consequent to the existence of
zero, there the set of all sets, and the law of the excluded middle,
with the equality operator, and other existential quantifiers, ie, for
any/each/all and exists.
Consider nonstandard analysis, NSA, it allows the consideration of
infinitesimals. In Cantor's time, he was very much against the notion
of the existence or even consideration of an infinitesimal, or rather
that's what I've read he said. Today we have a field of analysis
championed by Robinson that lets us consider infinitesimal numbers, as
well they can be considered as some of Conway's surreal numbers.
You might say that infinitesimals have naught to do with set theory,
but they do have a role in describing real numbers, which if set
theory is to be the foundation of all mathematics, is a set.
If we look to classical analysis, there are powerful tools for
determining empirical results, they're empirical in that we can
construct real-world experiments and verify them. For example, the
definite integral of many functions evaluates to equal the area within
a shape described by the graph or plot of that function, and analysis
is used every day to ensure that bridges and dams hold.
In the classical integral analysis, we have the tool of the Riemannian
limit, it allows us to sidestep the vagaries of infinitesimals which
were the foundation of the original founders of integral analysis'
infinitesimal analysis. Using today's nonstandard analysis, we can
verify those results of the classical analysis again with
infinitesimals.
There are perhaps also other tools that we might be able to use for
analysis. Just as non-Euclidean geometry allows elegant descriptions
that are either impossible to describe in the Euclidean geometry or
cumbersome, regardless holding true, if analysis of a bijective
mapping between the naturals and unit interval of reals holds true and
particularly if it provides a result unattainable from the opposite,
then a post-Cantorian set theory is a worthwhile consideration, valid
and sound, and thus part of mathematical knowledge.
May I interest you in a bridge?
Warm regards,
Ross Finlayson
Jesse F. Hughes
> je...@phiwumbda.org (Jesse F. Hughes) wrote in message news:<87u0uzb...@phiwumbda.org>...
>> r...@tiki-lounge.com (Ross A. Finlayson) writes:
>>
>> > Which do you prefer:
>> >
>> > super-Cantorian set theory
>> > non-Cantorian set theory
>> > post-Cantorian set theory
>> >
>> > Non-Euclidean geometry is also called extra- or super-Euclidean, or
>> > Minkowskian.
>>
>> Well, what theory do you have in mind and I'll tell you what I call
>> it.
>>
>> Write down some axioms. Non-Euclidean geometry didn't come about by
>> the handwaving speculation and diatribes you've offered. It came
>> about when folks negated the fifth postulate in one of two explicit
>> ways. Now which part of ZF do you intend to change and how?
>
> Hi,
>
> That's a good question.
>
> I think the axioms of ZF minus powerset and regularity are theorems of
> a set of zero axioms, where they are consequent to the existence of
> zero, there the set of all sets, and the law of the excluded middle,
> with the equality operator, and other existential quantifiers, ie, for
> any/each/all and exists.
You're blathering again. Pure and utter nonsense.
> May I interest you in a bridge?
You can try.
--
"A recruitment consultant I know thinks the most important quality in
a winner is to be lucky. To avoid wasting his time with unlucky
applicants, he takes half the resumes piled on his desk and throws
them straight in the bin." -- John Ramsden
Shmuel (Seymour J.) Metz
at 05:46 PM, r...@tiki-lounge.com (Ross A. Finlayson) said:
>Non-Euclidean geometry is also called extra- or super-Euclidean, or
>Minkowskian.
That's a bunch of bulya.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT <http://patriot.net/~shmuel>
Unsolicited bulk E-mail subject to legal action. I reserve the
right to publicly post or ridicule any abusive E-mail. Reply to
domain Patriot dot net user shmuel+news to contact me. Do not
reply to spam...@library.lspace.org
Andrew Usher
> You wrote:
>
> ,----[ <6e197594.04081...@posting.google.com> ]
> | Does the construction of L require AC to begin with?
> |
> | To go back to a previous example, AC -> there is a non-measurable
> | subset of R.
> |
> | If V=L -> AC, then L must contain such a set.
> |
> | Clearly, adding more sets to V (and making AC false) can't eliminate
> | anything, so there must be a non-measurable regardless of AC, if L is
> | contained in V, right?
> `----
>
> I don't know diddly about the definitions, but I imagine that the
> definition of non-measurable includes some quantifiers. If we enlarge
> the universe of discourse, then a set which was previously
> non-measurable may well become measurable. Someone that knows a
> definition or two can confirm or deny this, but I'd reckon that a set
> is not measurable/non-measurable all by its lonesome, but only in some
> context, namely the universe.
The universe, in this case, is R. This is contained in L, and adding
more sets can't change the properties of R.
> In any case, it is just painfully obvious that V >= L. L is the
> collection of all sets which provably exist. Any model of ZF must
> contain every set which provably exists (well, perhaps with some of
> these sets identified). Let's call a model M of ZF /good/ if,
> whenever M |= s = t for any terms s and t, then also ZF |- s = t.
> Some model theorist can tell me what the right terminology is, but
> "good" will do for now. Clearly, if M is good, then L <= M. That is,
> then L is a submodel of M.
You have passed the limits of my knowledge here; this notation is
gibberish to me.
> V is clearly a good model of ZF. Therefore L is a submodel (not
> necessarily proper) of V.
>
> This is more detail than ought to be necessary. Every constructible
> set is a *set*. Therefore, L is contained in V. Duh.
L is not 'constructible' by the ordinary definition constructible =
determined by a finite proof. Since L contains all the ordinals, it
must be larger than any cardinal. This is much larger than the
(countable) set of constructible objects.
Also, the construction of L requires uncountably many steps, since it
uses transfinite induction. Hence I can't accept L is a valid concept
anyway.
I see no reason why it can't be said that L is 'too big'. It's just
obvious that any universe containing AC will be larger than an
equivalent one that does not.
Andrew Usher
Ross A. Finlayson
> In <3c6b9c1e.04081...@posting.google.com>, on 08/18/2004
> at 05:46 PM, r...@tiki-lounge.com (Ross A. Finlayson) said:
>
> >Non-Euclidean geometry is also called extra- or super-Euclidean, or
> >Minkowskian.
>
> That's a bunch of bulya.
http://www.cs.nyu.edu/pipermail/fom/2003-February/006233.html
Loewenheim-Skolem says there are models where Cantor's powerset
mapping theorem is or is not true. As well, Loewenheim-Skolem has
models where P(N) is countable, aka denumerable. Is this not true?
Where EF is a bijection between N and R[0,1), it's not the exact same
result that
1
S 1 dx = 1
0
That is to say, the sum of each infinitesimal (1) dx for x from zero
to infinity equals 1, but the evaluation
00
S EF(x) dx
n=0
over the naturals is equal to two.
That's a bunch of infinitesimals.
Is Loewenheim-Skolem nonsense? How do you explain Banach-Tarksi?
(Inelegant insult omitted.) I'll be happy to assist you in explaining
Banach-Tarski.
Ross
Virgil
r...@tiki-lounge.com (Ross A. Finlayson) wrote:
> Loewenheim-Skolem says there are models where Cantor's powerset
> mapping theorem is or is not true.
As far as I can tell there aren't many models in which Cantor's powerset
mapping theorem is neither.
Jesse F. Hughes
> je...@phiwumbda.org (Jesse F. Hughes) wrote in message news:<87llgd3...@phiwumbda.org>...
>
>> You wrote:
>>
>> ,----[ <6e197594.04081...@posting.google.com> ]
>> | Does the construction of L require AC to begin with?
>> |
>> | To go back to a previous example, AC -> there is a non-measurable
>> | subset of R.
>> |
>> | If V=L -> AC, then L must contain such a set.
>> |
>> | Clearly, adding more sets to V (and making AC false) can't eliminate
>> | anything, so there must be a non-measurable regardless of AC, if L is
>> | contained in V, right?
>> `----
>>
>> I don't know diddly about the definitions, but I imagine that the
>> definition of non-measurable includes some quantifiers. If we enlarge
>> the universe of discourse, then a set which was previously
>> non-measurable may well become measurable. Someone that knows a
>> definition or two can confirm or deny this, but I'd reckon that a set
>> is not measurable/non-measurable all by its lonesome, but only in some
>> context, namely the universe.
>
> The universe, in this case, is R. This is contained in L, and adding
> more sets can't change the properties of R.
Don't be too hasty here. The properties of R can certainly change,
depending on what one means by properties. If by property, you mean
all of the true statements involving R, then *of course* this may
change by adding more sets, because *some* of these statements involve
quantifiers.
A statement "for all sets X, blah blah blah R blah X," may very well
be true in L and not in a larger universe.
>
>> In any case, it is just painfully obvious that V >= L. L is the
>> collection of all sets which provably exist. Any model of ZF must
>> contain every set which provably exists (well, perhaps with some of
>> these sets identified). Let's call a model M of ZF /good/ if,
>> whenever M |= s = t for any terms s and t, then also ZF |- s = t.
>> Some model theorist can tell me what the right terminology is, but
>> "good" will do for now. Clearly, if M is good, then L <= M. That is,
>> then L is a submodel of M.
>
> You have passed the limits of my knowledge here; this notation is
> gibberish to me.
Darn shame. Probably doesn't bode well for your ability to converse
sensibly on the relationship between V and L.
>
>> V is clearly a good model of ZF. Therefore L is a submodel (not
>> necessarily proper) of V.
>>
>> This is more detail than ought to be necessary. Every constructible
>> set is a *set*. Therefore, L is contained in V. Duh.
>
> L is not 'constructible' by the ordinary definition constructible =
> determined by a finite proof.
I didn't say L is a set. I said that every element of L is a
constructible set.
> Since L contains all the ordinals, it must be larger than any
> cardinal. This is much larger than the (countable) set of
> constructible objects. Also, the construction of L requires
> uncountably many steps, since it uses transfinite induction. Hence I
> can't accept L is a valid concept anyway.
>
> I see no reason why it can't be said that L is 'too big'. It's just
> obvious that any universe containing AC will be larger than an
> equivalent one that does not.
What does equivalent mean? How is V equivalent to L?
Every constructible set is a set. L cannot be larger than V. This is
just obvious.
--
Jesse F. Hughes
"Of course, my ability to admit my mistakes and correct them is a
trait that many of you seem to never have properly appreciated."
-- JSH, discussing his 1463rd "proof" of Fermat's Last Theorem.
Jesse F. Hughes
> "Shmuel (Seymour J.) Metz" <spam...@library.lspace.org.invalid> wrote in message news:<4124f4fa$3$fuzhry+tra$mr2...@news.patriot.net>...
>> In <3c6b9c1e.04081...@posting.google.com>, on 08/18/2004
>> at 05:46 PM, r...@tiki-lounge.com (Ross A. Finlayson) said:
>>
>> >Non-Euclidean geometry is also called extra- or super-Euclidean, or
>> >Minkowskian.
>>
>> That's a bunch of bulya.
>
> http://www.cs.nyu.edu/pipermail/fom/2003-February/006233.html
>
> Loewenheim-Skolem says there are models where Cantor's powerset
> mapping theorem is or is not true. As well, Loewenheim-Skolem has
> models where P(N) is countable, aka denumerable. Is this not true?
Every model M of set theory satisfies |N| < |P(N)|. Even the
countable models.
But you really need an introduction to model theory to understand how
we can have |M| = |N| and also M |= |N| < |P(N)|. The
internal/external distinction only leads to confusion without an
understanding of basic model theory.
Learn some logic *before* you try to claim that Cantor's theorem is
optional.
--
Jesse Hughes
"Surround sound is going to be increasingly important in future
offices."
-- Microsoft marketing manager displays his keen insight
Shmuel (Seymour J.) Metz
at 12:07 AM, r...@tiki-lounge.com (Ross A. Finlayson) said:
>http://www.cs.nyu.edu/pipermail/fom/2003-February/006233.html
That's nice. What does it have to do with anything I wrote? Did Bolya
plagiarize from Cantor? Did Gauss use a diagonal argument? Inquiring
minds want to know.
>Is this not true?
Is it not irrelevant to Non-Euclidean Geometry?
>Is Loewenheim-Skolem nonsense?
No, but your article is. Are you going to stop beating your wife?
>How do you explain Banach-Tarksi?
To whom? Euclid?
>I'll be happy to assist you in explaining Banach-Tarski.
Thank you, but were I to need assistance I'd ask someone who knows how
to read. I suggest that you read the text that you quoted from
<3c6b9c1e.04081...@posting.google.com> instead of shooting
from the hip; I quoted, and commented on, one sentence, and none of
your attempts at sarcasm have anything to do with that one sentence.
And BTW, the misspelling of Bolya is what is called a pun, which is a
form of humor[1].HTH. HAND. FOAD.
[1] Look it up in a dictionary if you are not familiar with the
concept.
Ross A. Finlayson
> In <3c6b9c1e.04081...@posting.google.com>, on 08/20/2004
> at 12:07 AM, r...@tiki-lounge.com (Ross A. Finlayson) said:
>
> >http://www.cs.nyu.edu/pipermail/fom/2003-February/006233.html
>
> That's nice. What does it have to do with anything I wrote? Did Bolya
> plagiarize from Cantor? Did Gauss use a diagonal argument? Inquiring
> minds want to know.
>
> >Is this not true?
>
> Is it not irrelevant to Non-Euclidean Geometry?
>
> >Is Loewenheim-Skolem nonsense?
>
> No, but your article is. Are you going to stop beating your wife?
>
> >How do you explain Banach-Tarksi?
>
> To whom? Euclid?
>
> >I'll be happy to assist you in explaining Banach-Tarski.
>
> Thank you, but were I to need assistance I'd ask someone who knows how
> to read. I suggest that you read the text that you quoted from
> <3c6b9c1e.04081...@posting.google.com> instead of shooting
> from the hip; I quoted, and commented on, one sentence, and none of
> your attempts at sarcasm have anything to do with that one sentence.
> And BTW, the misspelling of Bolya is what is called a pun, which is a
> form of humor[1].HTH. HAND. FOAD.
>
> [1] Look it up in a dictionary if you are not familiar with the
> concept.
Hi Shmuel,
It's nice of you to resort to a word that doesn't exist except of your
own construction if you don't want it to be understood. Thanks for
hinting that it was a pun on Bolyai, because otherwise it means
nothing. I had not heard too much of that fellow, you seem to refer
to Janos Bolyai, who independently developed some aspects of
hyperbolic (non-Euclidean) geometry in the milieu of Gauss, where
Gauss himself had his own notions on the matter of non-Euclidean
geometries. Both Gauss and Bolyai used the infinitesimal analysis.
http://www.google.com/search?hl=en&ie=ISO-8859-1&q=Janos+Bolya
So, by your pun, are you trying to imply that you've heard of someone
else much farther along in this line of argument, or that they never
heard of me, or you, or vice versa? Have you? Please let me know if
there is a rabid camp of Skolemites, and if so how to contact them.
This is mathematics: humor is irrelevant, although often appreciated,
and sarcasm is a lie. Do you think you're funny? It may be rude of
you to say that others aren't.
I want to tell you that you have some spelling errors on your web
page, "abd" and "expereience". That's not relevant to this argument,
syntax errors, but you might want to correct them. Your misspellings
are what are called errors. When you wrote "bulya", I thought it was
a correctly-spelled Hebrew word, because I'm naive and trusting.
Humor is basically an implicit or explicit truth. Then again, some
people have a sixth sense of humor. I generally don't use sarcasm, as
you might know from many postings to sci.math about humor. Now, that
hasn't been too fair because you've probably said something that was
funny in your life, it would be rude of me to suggest otherwise.
Shmuel, 2+2=4, and in a model of ubiquitous ordinals, infinite sets
are equivalent.
Loewenheim-Skolem says what I said it does above.
Have your theory: in a post-Cantorian theory you are allowed to
consider a hypothesis where infinite set maps, 1-1, onto, and
bijectively, to powerset. Then you can forget about it.
http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Lowenheim.html
"Loewenheim is remembered for the Loewenheim-Skolem paradox (which
Skolem pointed out is not a paradox!) which produces non-standard
models, for example a denumerable model of the reals."
Translation: Loewenheim says the reals are countable. I wish I'd
known about this before, I could have invoked Loewenheim-Skolem
instead of independently developing appropriate notions myself, for
several of the contentious issues we discuss. Now I won't deny that
they say so, unless necessary.
EF is the primitive function that meets requirements to be a bijective
mapping between the naturals and reals, in extension of Cantor.
http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Skolem.html
Now those are some logicians I appreciate: debunking paradoxes.
So, if we're talking about Loewenheim and Skolem, we either agree or
don't that they might agree.
What do your acronyms mean there? Never mind...
Hey where's this from: "They laughed at me when I told them I'd
become a great comedian. Nobody laughs at me now."
Good luck finding someone literate.
Shmuel (Seymour J.) Metz
at 02:09 PM, r...@tiki-lounge.com (Ross A. Finlayson) said:
>It's nice of you to resort to a word that doesn't exist except of
>your own construction if you don't want it to be understood.
Well, most people who failed to get the joke would have assumed that
it was a typo and proceded from there.
>Janos Bolyai, who independently developed some aspects of hyperbolic
>(non-Euclidean) geometry
There you go again.
>So, by your pun, are you trying to imply that you've heard of
>someone else much farther along in this line of argument,
No, I was trying to humorously convey the basic concept that
Hyperbolic Geometry is not the same as Non-Euclidean Geometry, and
that Bolyai worked on a non-Euclidean Geometry that was *NOT*
Hyperbolic.
>rabid camp of Skolemites,
Not relevant to the difference between elliptical and hyperbolic.
>I want to tell you that you have some spelling errors on your web
>page, "abd" and "expereience".
Thanks you. I've corrected those.
>I generally don't use sarcasm
But you did in this case. Inappropriately, since you had missed, and
continued to miss, the entire point.
>Shmuel, 2+2=4, and in a model of ubiquitous ordinals, infinite sets
>are equivalent.
What does that have to do with the point at issue?
>Loewenheim-Skolem says what I said it does above.
What does that have to do with the point at issue?
>So, if we're talking about Loewenheim and Skolem,
No. The text that I quoted and commented on had nothing to do with
Loewenheim and Skolem, as I already pointed out to you. But perhaps
you're only pretending to be dense as a form of humor and I'm simply
failing to understand how funny it is. Or maybe "Non-Euclidean
geometry is also called extra- or super-Euclidean, or Minkowskian."
was intended as a joke.
Andrew Usher
> >> ,----[ <6e197594.04081...@posting.google.com> ]
> >> | Does the construction of L require AC to begin with?
> >> |
> >> | To go back to a previous example, AC -> there is a non-measurable
> >> | subset of R.
> >> |
> >> | If V=L -> AC, then L must contain such a set.
> >> |
> >> | Clearly, adding more sets to V (and making AC false) can't eliminate
> >> | anything, so there must be a non-measurable regardless of AC, if L is
> >> | contained in V, right?
> >> `----
> >>
> >> I don't know diddly about the definitions, but I imagine that the
> >> definition of non-measurable includes some quantifiers. If we enlarge
> >> the universe of discourse, then a set which was previously
> >> non-measurable may well become measurable. Someone that knows a
> >> definition or two can confirm or deny this, but I'd reckon that a set
> >> is not measurable/non-measurable all by its lonesome, but only in some
> >> context, namely the universe.
> >
> > The universe, in this case, is R. This is contained in L, and adding
> > more sets can't change the properties of R.
>
> Don't be too hasty here. The properties of R can certainly change,
> depending on what one means by properties. If by property, you mean
> all of the true statements involving R, then *of course* this may
> change by adding more sets, because *some* of these statements involve
> quantifiers.
The real numbers are a definite entity. If two theories predict
differing properties of R, one must be wrong. If Lesbesgue measure has
quantifiers in this sense, then it is inconsistent.
> >> In any case, it is just painfully obvious that V >= L. L is the
> >> collection of all sets which provably exist. Any model of ZF must
> >> contain every set which provably exists (well, perhaps with some of
> >> these sets identified). Let's call a model M of ZF /good/ if,
> >> whenever M |= s = t for any terms s and t, then also ZF |- s = t.
> >> Some model theorist can tell me what the right terminology is, but
> >> "good" will do for now. Clearly, if M is good, then L <= M. That is,
> >> then L is a submodel of M.
> >
> > You have passed the limits of my knowledge here; this notation is
> > gibberish to me.
>
> Darn shame. Probably doesn't bode well for your ability to converse
> sensibly on the relationship between V and L.
I am not a crank! Perhaps direct me to an explanation I could
understand of how L is constructed; I have examined texts on set
theory but find them either to be too simple to discuss this, or
incomprehensible to me.
> >> V is clearly a good model of ZF. Therefore L is a submodel (not
> >> necessarily proper) of V.
> >>
> >> This is more detail than ought to be necessary. Every constructible
> >> set is a *set*. Therefore, L is contained in V. Duh.
> >
> > L is not 'constructible' by the ordinary definition constructible =
> > determined by a finite proof.
>
> I didn't say L is a set. I said that every element of L is a
> constructible set.
Sorry, I meant 'every element of L'. There are only a countable number
of constructible sets, therefore some elements of L are not
constructible.
> > Since L contains all the ordinals, it must be larger than any
> > cardinal. This is much larger than the (countable) set of
> > constructible objects. Also, the construction of L requires
> > uncountably many steps, since it uses transfinite induction. Hence I
> > can't accept L is a valid concept anyway.
> >
> > I see no reason why it can't be said that L is 'too big'. It's just
> > obvious that any universe containing AC will be larger than an
> > equivalent one that does not.
>
> What does equivalent mean? How is V equivalent to L?
Since AC does not imply V=L, this is a non sequitur.
> Every constructible set is a set. L cannot be larger than V. This is
> just obvious.
No, that is meaningless. For an arbitrary element of L, you can't give
me a construction for it. The sense in which L is 'constructible' is a
different meaning of the word 'constructible' than the one used in
stating that it is 'obvious' that constructible sets all exist.
There are two kinds of mathematics: the useful, meaningful part
dealing with finite and countable sets, computable numbers, and
constructible objects; and the division dealing with uncountable sets,
undefinable objects, and unintelligible concepts - nonsense upon
nonsense upon nonsense. You apparently belong to the latter school.
Andrew Usher
Jesse F. Hughes
>> Don't be too hasty here. The properties of R can certainly change,
>> depending on what one means by properties. If by property, you mean
>> all of the true statements involving R, then *of course* this may
>> change by adding more sets, because *some* of these statements involve
>> quantifiers.
>
> The real numbers are a definite entity. If two theories predict
> differing properties of R, one must be wrong. If Lesbesgue measure has
> quantifiers in this sense, then it is inconsistent.
The "properties" are different only if we understand "property" to be
synonymous with "formal statement", and our interpretation of certain
formal statements has changed.
Consider a statement about R of the form "For all sets X, blah(R,X)".
If we call such a statement a property, then *of course* whether R
satisfies that statement can depend on what sets we have in our
universe.
Statements like "R is Cauchy-complete" are exactly this kind of
statement. What Cauchy-completeness means depends on what Cauchy
sequences are available.
Given L c V, it's not obvious that the set R in L is the same set as
the set R in V. If there are more sets in V, then there may be more
Cauchy sequences in V than in L and so the unique set satisfying the
definition of R in V may be a superset of the unique set satisfying
the definition of R in L.
I don't know if that's the case. I don't study these issues.
However, the idea that the reals are a definite entity independent of
what set constructions are available to us seems a bit naive.
--
Jesse F. Hughes
"Yesterday was Judgment Day. How'd you do?"
-- The Flatlanders
Nath Rao
> "Loewenheim is remembered for the Loewenheim-Skolem paradox (which
> Skolem pointed out is not a paradox!) which produces non-standard
> models, for example a denumerable model of the reals."
>
> Translation: Loewenheim says the reals are countable. I wish I'd
> known about this before, I could have invoked Loewenheim-Skolem
> instead of independently developing appropriate notions myself, for
> several of the contentious issues we discuss. Now I won't deny that
> they say so, unless necessary.
You have misunderstood Lowenheim-Skolem.
It says that any consistent first-order theory with a countable number
of constants and axioms has a countable model. This means that the model
is countable in the >meta<-theory. A 'countable' model of ZFC just means
that there is some relation R on natural numbers, which when interpreted
as membership, satisfies the ZFC axioms. But no function (in the
meta-theory) from the 'reals' (of the model) to the naturals of the
meta-theory will be a set of the model (obviously). You cannot count the
reals of the model >within< the model. The relation R satisfies no
useful property, and its existence is quite mysterious as it is proved
by contradiction.
Anyway, the definition of reals implicitly used by mathematicians is
second order.
David McAnally
>k_over...@yahoo.com (Andrew Usher) writes:
>>> Don't be too hasty here. The properties of R can certainly change,
>>> depending on what one means by properties. If by property, you mean
>>> all of the true statements involving R, then *of course* this may
>>> change by adding more sets, because *some* of these statements involve
>>> quantifiers.
>>
>> The real numbers are a definite entity. If two theories predict
>> differing properties of R, one must be wrong. If Lesbesgue measure has
>> quantifiers in this sense, then it is inconsistent.
>The "properties" are different only if we understand "property" to be
>synonymous with "formal statement", and our interpretation of certain
>formal statements has changed.
>Consider a statement about R of the form "For all sets X, blah(R,X)".
>If we call such a statement a property, then *of course* whether R
>satisfies that statement can depend on what sets we have in our
>universe.
>Statements like "R is Cauchy-complete" are exactly this kind of
>statement. What Cauchy-completeness means depends on what Cauchy
>sequences are available.
>Given L c V, it's not obvious that the set R in L is the same set as
>the set R in V.
Typically, it isn't.
>If there are more sets in V, then there may be more
>Cauchy sequences in V than in L and so the unique set satisfying the
>definition of R in V may be a superset of the unique set satisfying
>the definition of R in L.
>I don't know if that's the case. I don't study these issues.
It is the case.
>However, the idea that the reals are a definite entity independent of
>what set constructions are available to us seems a bit naive.
In any model of ZFC, R is completely determined up to isomorphism. On the
other hand, given any model V of ZFC, and any two infinite sets A and B in
V, there exists a generic extension V[G] of V in which A and B are
bijective. Specifically, there is a generic extension V[G] in which N and
R^V (the realization of R in V) are bijective, i.e. there is a generic
extension V[G] of V in which R^V is countable. Obviously, in such an
extension, R^V cannot be regarded as a model for the real numbers in V[G],
even though it is a model for the real numbers in V. In the generic
extension, there are Cauchy sequences which do not occur in V (there are,
in V[G], Cauchy sequences in R^V which do not have a limit in R^V - such
Cauchy sequences are elements of V{G}, but not elements of V). It follows
that there are generic extensions of L in which there are real numbers
which are not constructible.
David
-----
Jesse F. Hughes
> je...@phiwumbda.org (Jesse F. Hughes) writes:
>
>>I don't know if that's the case. I don't study these issues.
>
> It is the case.
Thanks for giving a better informed opinion on this.
--
Jesse Hughes
"Like the ski resort full of girls hunting for husbands
and husbands hunting for girls, the situation is not as
symmetrical as it might seem." -- Alan MacKay
Ross A. Finlayson
Hi Nath,
I necessarily disagree.
Instead, I theorize that ordinals are ubiquitous, and any function
between the set of natural numbers and the set that is the set of real
numbers is an implicit composition with EF.
That's the problem with demanding a metatheory, instead of just having
a theory, that as the variable of the n'th order logic n diverges,
it's still subject to the same problems of considering an infinite
ordinal.
That reminds me of hearing about "epsilon chains of length sixteen" or
"eleven dimensions". The universe is n-dimensional.
If these are nonstandard, let's consider a nonstandard model of the
reals: the hyperreals. The reals are just the hyperreals, and vice
versa, and the integers are just the hyperintegers, and vice versa.
They're just called nonstandard so they can be discussed rationally.
The second order logic, and n'th order logic, can only be within a
first order logic.
EF is a function between the set N and the unit interval of the set R.
"No classes in set theory, and no models in theory."
Ross F.
Herman Jurjus
[snip]
> In any model of ZFC, R is completely determined up to isomorphism. On the
> other hand, given any model V of ZFC, and any two infinite sets A and B in
> V, there exists a generic extension V[G] of V in which A and B are
> bijective.
Is this true? Do you have any references, or a quick argument?
Because if it is true, doesn't this mean that Finlayson doesn't talk
complete nonsense, when he insists that all infinite sets are alike?
--
Cheers,
Herman Jurjus
Jesse F. Hughes
Well, I wouldn't say that.
(*) R is never equinumerous (internally) with N in a given model M.
(**) The set R-in-M is equinumerous (internally) with N-in-M in a
generic extension M' of M.
That's not much alike.
(By the way, I don't know David's argument or even the meaning of
"generic extension". I'm just taking the last question about Ross not
talking complete nonsense. It's the easiest.)
--
"Not all features that are found on the Security tab are designed to
help make your documents and files more secure." --Microsoft on Office
security features (after it was pointed out by a third party that a
certain password setting is easily bypassed.)
David McAnally
>David McAnally wrote:
>[snip]
>> In any model of ZFC, R is completely determined up to isomorphism. On the
>> other hand, given any model V of ZFC, and any two infinite sets A and B in
>> V, there exists a generic extension V[G] of V in which A and B are
>> bijective.
>Is this true?
Yes.
>Do you have any references, or a quick argument?
Let the set of forcing conditions be the set of bijections between finite
subsets of A and finite subsets of B. For two conditions p and q, q is
stronger than p if q contains p as a subset. This means that p and q are
compatible iff p(x) = q(x) for all x in the intersection of dom(p) and
dom(q), and p^{-1}(y) = q^{-1}(y) for all y in the intersection of
range(p) and range(q). The union of a generic set G is a bijection
between A and B in V[G].
>Because if it is true, doesn't this mean that Finlayson doesn't talk
>complete nonsense, when he insists that all infinite sets are alike?
Not at all. In any model of ZFC, there is more than one infinite
cardinality (the class of infinite cardinalities is actually a proper
class). So not all infinite sets are alike. Specifically, in any model of
ZFC, R is uncountable, so R and N are not bijective. The closest that the
claim that all infinite sets are alike comes to the truth is the fact that
there are countable models of ZFC, and in such a model, all infinite sets
are represented by countable sets. The bijections between such countable
sets are not generally themselves represented by elements of the model.
David
And all dared to brave unknown terrors, to do mighty deeds,
to boldly split infinitives that no man had split before -
and thus was the Empire forged.
-----
Herman Jurjus
> Herman Jurjus <h.ju...@hetnet.nl> writes:
>
>
>>David McAnally wrote:
>>[snip]
>
>
>>>In any model of ZFC, R is completely determined up to isomorphism. On the
>>>other hand, given any model V of ZFC, and any two infinite sets A and B in
>>>V, there exists a generic extension V[G] of V in which A and B are
>>>bijective.
>
>
>>Is this true?
>
>
> Yes.
>
>
>>Do you have any references, or a quick argument?
>
>
> Let the set of forcing conditions be the set of bijections between finite
> subsets of A and finite subsets of B. For two conditions p and q, q is
> stronger than p if q contains p as a subset. This means that p and q are
> compatible iff p(x) = q(x) for all x in the intersection of dom(p) and
> dom(q), and p^{-1}(y) = q^{-1}(y) for all y in the intersection of
> range(p) and range(q). The union of a generic set G is a bijection
> between A and B in V[G].
But surely this only makes sense if V is countable? What if V is
uncountable,
and R 'in' V is also uncountable (also when looked at 'from the outside'
so to say), then surely you can't extend V and cook up some bijection
with N? That *would* be rather shocking.
--
Cheers,
Herman Jurjus
David McAnally
As I see it, the model of ZFC is represented by a nonempty set (so V is a
set as seen from the outside, but not a set in the model). The 'outside'
is itself a model of ZFC, and it may be possible that the 'outside' can
itself be extended generically to a model in which V is countable, and in
such an extended model of the 'outside', there is no need to worry. Of
course, this will take more thought than these initial musings. For
example, thinking about the size of the language, and the number of
constant symbols, etc, may lead to other considerations.
David
-----
Ross A. Finlayson
Hi David,
Progress!
You agree that "there are countable models of ZFC, and in such a
model, all infinite sets are represented by countable sets"? I'm
wary.
Are not the powerset of the set of naturals or the set of reals sets
in those "models"? Otherwise there must be some other justification
of your refusal to accept any post-Cantorian logic. The set of reals
is a set in ZFC, is that not so? How about ZF?
What do you think about the notion that there can only be one or no
proper classes? What do you think about a theory where some value,
say, the proper class, is dually represented as zero and infinity?
I have plentiful self-effacing humor. It does not support my effort
to call others' nonsense without gently cajoling, nor to overreact,
because I do not push with the weight of establishment, only with
reason. The fallacy of "Not at all" is trivial.
For some, an introduction to Cohen's forcing and Loewenheim and
Skolem's theorems in this thread have changed their fundamental
perceptions.
I speak and write rightly and honestly, and with sense.
Regards,
Ross F.
Jesse F. Hughes
>
> Hi David,
>
> Progress!
>
> You agree that "there are countable models of ZFC, and in such a
> model, all infinite sets are represented by countable sets"? I'm
> wary.
Learn some model theory, so that you can understand that this result
has nothing at all to do with your "post-Cantorian" blathering.
> Are not the powerset of the set of naturals or the set of reals sets
> in those "models"? Otherwise there must be some other justification
> of your refusal to accept any post-Cantorian logic. The set of reals
> is a set in ZFC, is that not so? How about ZF?
What "post-Cantorian logic"? You haven't shown us any.
Want to do set theory in which there is only one infinite cardinal?
Okay. I won't say it's impossible, but you'll have to show me the
axioms that support this aim. How do they differ from ZF? In ZF, it
is a plain fact that for every X, there is no surjection X -> PX.
The L-S theorem does not change that fact.
> I have plentiful self-effacing humor.
Few people have greater needs for it.
> For some, an introduction to Cohen's forcing and Loewenheim and
> Skolem's theorems in this thread have changed their fundamental
> perceptions.
For whom? You? You haven't been introduced to L-S. You really must
work through the introductory model theory material that comes before
it in order to understand that the L-S theorem is not the pivot for
moving the universe away from Cantor's theorem.
> I speak and write rightly and honestly, and with sense.
Is this that self-effacing humor again?
--
Jesse Hughes
"I often told you of the dangers of hubris, and most importantly of
all, I TOLD you that I wanted to change the institution of mathematics
worldwide." -- James Harris, on the evils of pride
Ross A. Finlayson
That's pretty funny. Not if you're me, but for you, it's funny.
I have a theory where the powerset is the order type is the successor,
for finite or infinite ordinals.
You'll notice that in transfinite cardinal arithmetic, N+1 = N, in my
theory, N+1 = P(N). In my theory, 1+1 = 2, and 2 = P(1) = S(1), S for
successor, P for powerset.
Are you missing an S in your powerset mapping for infinite sets?
That's pretty funny, for me.
Are not the naturals or reals sets in those "models"? Unlearn some
model theory.
It's L-S all the way out!
I looked at the definition of "proper class" and it precludes there
being more than one.
Unitize the analog! The integral bar is a stylized summation sign.
It signifies summing the integrand for each value of the domain, where
you'll notice there is that "little" dx in, or near, the integrand.
Recently I saw several others claim that lim n->oo sinc(x) = 0.
That's progress.
Progress!
Good luck,
Ross
David McAnally
>Herman Jurjus <h.ju...@hetnet.nl> writes:
>> David McAnally wrote:
>> [snip]
>>
>>> In any model of ZFC, R is completely determined up to isomorphism. On the
>>> other hand, given any model V of ZFC, and any two infinite sets A and B in
>>> V, there exists a generic extension V[G] of V in which A and B are
>>> bijective.
>>
>> Is this true? Do you have any references, or a quick argument?
>>
>> Because if it is true, doesn't this mean that Finlayson doesn't talk
>> complete nonsense, when he insists that all infinite sets are alike?
>Well, I wouldn't say that.
>(*) R is never equinumerous (internally) with N in a given model M.
>(**) The set R-in-M is equinumerous (internally) with N-in-M in a
>generic extension M' of M.
>That's not much alike.
>(By the way, I don't know David's argument or even the meaning of
>"generic extension". I'm just taking the last question about Ross not
>talking complete nonsense. It's the easiest.)
For myself, the easiest way to understand generic extensions is through
Boolean-valued models.
Let V be a model for ZFC, and let B be a complete Boolean algebra (so that
B is an element of V, {((x,y),x+y) : x, y in B} is an element of V,
{((x,y),x.y) : x, y in B} is an element of V, {(x,x') : x in B} is an
element of V, {(A,sup(A)) : A in V, A a subset of B} is an element of V,
and {(A,inf(A)) : A in V, A a subset of B} is an element of V). For all
ordinals a, define the element V^B_a by:
V^B_0 = empty set,
V^B_{a+1} = {x in V : x is a function, dom(x) is a subset of
V^B_a, and range(x) is a subset of B},
V^B_a = U_{b < a} V^B_b if a is a limit ordinal,
where U_{b < a} V^B_b denotes the union of V^B_b over all b < a.
V^B is the union of V^B_a over all ordinals a. Note that V^B_a is a set
in V for each ordinal a, but that V^B is a proper class (and a subclass
of V).
For x in V, we can define \hat x in V^B inductively by
dom(\hat x) = {\hat y : y in x},
\hat x(\hat y) = 1 for all y in x.
This gives an embedding of V in V^B as a subclass.
For x and y in V^B, ||x in y||, ||x subset y|| and ||x = y|| can be
defined inductively by
||x in y|| = sup{ y(z).||x = z|| : z in dom(y) },
||x subset y|| = inf{ x(z)'+||z in y|| : z in dom(x) },
||x = y|| = ||x subset y||.||y subset x||,
so that the "truth values" of "x in y", "x subset y" and "x = y" are given
as elements of B.
Specifically, it is demonstrable that
||\hat x in \hat y|| = 1 if x, y in V and x in y,
||\hat x in \hat y|| = 0 if x, y in V and x not in y,
||\hat x subset \hat y|| = 1 if x, y in V and x is a subset of y,
||\hat x subset \hat y|| = 0 if x, y in V and x is not a subset of y,
||\hat x = \hat y|| = 1 if x, y in V and x = y,
||\hat x = \hat y|| = 0 if x, y in V and x not equal to y,
||x = x|| = 1 for all x in V^B,
||x in x|| = 0 for all x in V^B.
For a formula \phi with assigned values in V^B for its free variables, we
can define ||\phi|| in B by induction on \phi by
||not \phi|| = ||\phi||',
||\phi and \psi|| = ||\phi||.||\psi||,
||\phi or \psi|| = ||\phi||+||\psi||,
||\phi implies \psi|| = ||\phi||'+||\psi||,
||for all x \phi(x)|| = inf{ ||\phi(x)|| : x in V^B },
||exists x \phi(x)|| = sup{ ||\phi(x)|| : x in V^B }.
Under these definitions, it is demonstrable that
||for all x in y \phi(x)|| = inf{ y(x)'+||\phi(x)|| : x in dom(y) },
||exists x in y \phi(x)|| = sup{ y(x).||\phi(x)|| : x in dom(y) }.
For any sentence \phi, if \phi is a theorem of ZFC, then ||\phi|| = 1.
It follows that if \phi is inconsistent with ZFC, then ||\phi|| = 0.
Since ||\phi|| = 1 if \phi is a theorem of ZFC, then V^B is a Boolean-
valued model of ZFC.
A generic ultrafilter G (not necessarily an element of V) is an
ultrafilter on B, so that
(1) every element of G is an element of B, but 0 is not an
element of G,
(2) if x is an element of G and x is less than or equal to y,
then y is an element of G,
(3) if x and y are elements of G, then x.y is an element of G,
(4) for any element x of B, either x is an element of G or x'
is an element of G (from (1) and (3), it follows that
exactly one of x and x' is an element of G).
As well as being an ultrafilter, a generic ultrafilter G also satisfies
(5) if A is an element of V and a subset of G, then inf(A) is
an element of G.
For all x in V^B, define i_G(x) by induction on x by
i_G(x) = { i_G(y) : y in dom(x), x(y) in G }.
Note that since G is not generally an element of V, then i_G(x) is not
generally an element of V. V[G] is the class with elements i_G(x) for
x in V^B.
For a sentence \phi, V[G] satisfies \phi iff ||\phi|| is an element of G.
It follows that V[G] is a model of ZFC, since ||\phi|| = 1 for all
theorems \phi of ZFC.
If \phi(x_1,...,x_k) is a formula in ZFC with free variables x_1,...,x_k,
then for elements y_1,...,y_k, in V^B,
V[G] satisfies \phi(i_G(y_1),....,i_G(y_k)) iff
||\phi(y_1,...,y_k)|| is an element of G.
For all x in V, i_G(\hat x) = x, so that V is a submodel of V[G].
Define G' in V^B by dom(G') = { \hat u : u in B } and G'(\hat u) = u for
all u in B. Then i_G(G') = { u in B : u in G } = G, and so G is an
element of V[G].
So V[G] is a model of ZFC that contains V as a submodel and G as an
element. Other properties include
(1) V[G] is the smallest model of ZFC that contains V as a
submodel and G as an element;
(2) The ordinals in V[G] are identical to the ordinals in V
(i.e. no new ordinals are introduced),
(3) The constructible sets in V[G] are the constructible sets
in V.
As an extra note about generic ultrafilters, ||G' is a generic
ultrafilter|| = 1, so that if ZFC is consistent, then the existence
of generic ultrafilters is consistent with ZFC.
David
-----
David McAnally
>je...@phiwumbda.org (Jesse F. Hughes) writes:
>>Herman Jurjus <h.ju...@hetnet.nl> writes:
>>> David McAnally wrote:
>>> [snip]
>>>
>>>> In any model of ZFC, R is completely determined up to isomorphism. On the
>>>> other hand, given any model V of ZFC, and any two infinite sets A and B in
>>>> V, there exists a generic extension V[G] of V in which A and B are
>>>> bijective.
>>>
>>> Is this true? Do you have any references, or a quick argument?
>>>
>>> Because if it is true, doesn't this mean that Finlayson doesn't talk
>>> complete nonsense, when he insists that all infinite sets are alike?
>>Well, I wouldn't say that.
>>(*) R is never equinumerous (internally) with N in a given model M.
>>(**) The set R-in-M is equinumerous (internally) with N-in-M in a
>>generic extension M' of M.
>>That's not much alike.
>>(By the way, I don't know David's argument or even the meaning of
>>"generic extension". I'm just taking the last question about Ross not
>>talking complete nonsense. It's the easiest.)
>For myself, the easiest way to understand generic extensions is through
>Boolean-valued models.
An alternative approach uses forcing conditions. Let F be an ordered
set in V. The elements of F are called conditions (or forcing
conditions), and the condition q is said to be stronger than the condition
p if q <= p (where q <= p denotes that q is less than or equal to p). Two
conditions p and q are said to be compatible if there is a condition r
which is stronger than both p and q, otherwise p and q are incompatible.
F can be endowed with a topology, with an open base given by { [p] : p in
F }, where [p] = { q in F : q <= p }. A subset A of F is open iff every
condition stronger than an element of A is also an element of A. A subset
A of F is closed iff every condition weaker than an element of A is also
an element of A. Since a dense subset D of F is a subset with nonempty
intersection with every nonempty open set, it follows that D is dense iff
for all conditions p there exists a condition q stronger than p such that
q is an element of D.
In a topological space, a regular open set is an open set which is the
interior of a closed set, so that an open set is regular iff it is the
interior of its closure. The regular open sets of a topological space
X form a complete Boolean algebra with
0 = empty set,
1 = X,
A+B = int(cl(A u B)),
A.B = A n B,
A' = int(X-A),
sup(M) = int(cl(union of M)) for M nonempty,
inf(M) = int(cl(intersection of M)) for M nonempty,
A <= B iff A is a subset of B,
where for sets S and T, S u T denotes the union of S and T, S n T denotes
the intersection of S and T, cl(S) denotes the closure of S, and int(S)
denotes the interior of S.
In the case of the topological space of forcing conditions F, denote
int(cl([p])) by e(p), so that e(p) is a regular open set. e(p) is the set
of conditions q such that every condition which is compatible with q is
also compatible with p, i.e. e(p) is the set of all conditions q such that
every condition which is incompatible with p is also incompatible with q.
The complete Boolean algebra RO(F) of regular open sets can now be used to
generate a Boolean-valued model of ZFC. If A is a regular open set and p
is an element of A, then e(p) is a subset of A. So, for any formula
\phi(x_1,...,x_k) with free variables x_1,...,x_k, and elements y_1,...,
y_k, of V^B (B = RO(F)), a condition p is an element of ||\phi(y_1,...,y_k)||
iff e(p) is subset of ||\phi(y_1,...,y_k)||, and p is said to force
\phi(y_1,...,y_k) under these equivalent conditions. Then
p forces not \phi iff for all q <= p, q does not force \phi;
if p forces \phi and q <= p, then q forces \phi;
p forces \phi and \psi iff p forces \phi and p forces \psi;
p forces \phi or \psi iff for all q <= p, there exists r <= q such that
r forces \phi or r forces \psi;
p forces for all x \phi(x) iff p forces \phi(x) for all x in V^B;
p forces exists x \phi(x) iff for all q <= p, there exist x in V^B and
r <= q such that r forces \phi(x).
A set G (not necessarily an element of V) is called a generic set if
(1) G is a subset of F (i.e. all the elements of G are forcing
conditions),
(2) if p is an element of G, then any condition q weaker than p
is an element of G,
(3) if p and q are elements of G, then p and q are compatible,
(4) if D is an element of V and D is a dense set of conditions,
then G and D have nonempty intersection, i.e. G intersects
every dense subset of F.
For all x in V^B, define i_G(x) by induction on x by
i_G(x) = { i_G(y) : y in dom(x), G intersects x(y) },
and define V[G] to be the class with elements i_G(x) for x in V^B.
Then V[G] is the smallest model of ZFC which contains V as a submodel and
contains G as an element. V[G] has the same ordinals as V, and the same
constructible sets as V.
For a formula \phi(x_1,...,x_k) with free variables x_1,...,x_k, and for
elements y_1,...,y_k, of V^B, V[G] satisfies \phi(i_G(y_1),...,i_G(y_k))
iff there exists a condition p in G such that p forces \phi(y_1,...,y_k).
Define G' in V^B by dom(G') = { \hat p : p in F }, and G'(p) = e(p) for
all conditions p in F, then ||G' is a generic set|| = 1, and i_G(G') = G.
It follows that if ZFC is consistent, then the existence of a generic set
is consistent with ZFC.
The connection between generic ultrafilters and generic sets is given as
follows.
If G is a generic set, then
U = { A a regular open set in F : A intersects G }
is the corresponding generic ultrafilter, and V[U] = V[G].
If U is a generic ultrafilter, then
G = { p in F : e(p) in U }
is the corresponding generic set, and V[G] = V[U].
Now, suppose that V is a model of ZFC, and let A and B be infinite sets
in V. Let the forcing conditions be bijections between finite subsets of
A and finite subsets of B, and let the forcing conditions be ordered by
inverse inclusions (i.e. q <= p if p is a subset of q), then ||there
exists a bijection between \hat A and \hat B|| = 1, and for any generic
set G, the union of G is a bijection between A and B in V[G].
As another example, in ZFC, the cardinality of R is a transfinite cardinal
not cofinal with omega. Conversely, for any transfinite cardinal not
cofinal with omega, there is a generic extension L[G] of L such that the
cardinality of R is the given cardinal.
David
-----
Jesse F. Hughes
>
> I have a theory where the powerset is the order type is the successor,
> for finite or infinite ordinals.
What does it mean, "you have a theory"?
What is the theory? What are the axioms?
>
> You'll notice that in transfinite cardinal arithmetic, N+1 = N, in my
> theory, N+1 = P(N). In my theory, 1+1 = 2, and 2 = P(1) = S(1), S for
> successor, P for powerset.
>
> Are you missing an S in your powerset mapping for infinite sets?
>
> That's pretty funny, for me.
>
> Are not the naturals or reals sets in those "models"? Unlearn some
> model theory.
Show me some axioms.
> It's L-S all the way out!
Lowenheim-Skolem does nothing to justify your delusions.
> I looked at the definition of "proper class" and it precludes there
> being more than one.
Which definition precludes that? Not the standard meaning of the
term.
>
> Unitize the analog! The integral bar is a stylized summation sign.
> It signifies summing the integrand for each value of the domain, where
> you'll notice there is that "little" dx in, or near, the integrand.
>
> Recently I saw several others claim that lim n->oo sinc(x) = 0.
> That's progress.
>
> Progress!
Blathering!
--
"It has been shown that no man can sit down to write without a very profound
design. Thus to authors in general trouble is spared. A novelist, for example,
need have no care of his moral. It is there -- that is to say, it is somewhere
-- and the moral and the critics can take care of themselves." --E.A. Poe
Ross A. Finlayson
> r...@tiki-lounge.com (Ross A. Finlayson) writes:
>
> >
> > I have a theory where the powerset is the order type is the successor,
> > for finite or infinite ordinals.
>
>
> What is the theory? What are the axioms?
>
There are no axioms, that way it can be complete in extension of
Goedel. It just uses zero/empty, tautology, and excluded middle,
paraconsistently.
That's what I've said for years.
You'll notice that N U {N} = N+1 = |N| = P(N) = succ(N), and that each
powerset of N is not a limit ordinal, and that the only limit ordinal
is N and 0.
You can read about the proper class consideration in "Ubiquitous
Naturals, Infinity".
Read!
Ross
KRamsay
In article <6e197594.04082...@posting.google.com>,
k_over...@yahoo.com (Andrew Usher) writes:
|There are two kinds of mathematics: the useful, meaningful part
|dealing with finite and countable sets, computable numbers, and
|constructible objects; and the division dealing with uncountable sets,
|undefinable objects, and unintelligible concepts - nonsense upon
|nonsense upon nonsense. You apparently belong to the latter school.
If you look at what people count as "applied math", quite a lot of it
is real analysis, dealing with R, the reals, an uncountable set. Are
you saying real analysis is unintelligible nonsense, or what?
Keith Ramsay
KRamsay
In article <2ogm7kF...@uni-berlin.de>, Herman Jurjus <h.ju...@hetnet.nl>
writes:
|Coming back to the original question: in (some forms of) intuitionist
|mathematics, N has uncountable subsets, so 'larger size' has very little
|to do with uncountability.
I hope you're not confusing constructive mathematics with intuitionist
mathematics. Intuitionism is the term Brouwer used to describe his
own philosophy of mathematics, and it's a more specific term. The
Markov school of constructive mathematics allegedly made free use
of the assumption that each function from the natural numbers N to
N is recursive (or equivalently Turing computable), which would imply
the existence of uncountable subsets of N. But I don't think they are
called intuitionists. It may be that some of Brouwer's assumptions,
or those of schools following him, also imply the existence of an
uncountable subset of N, but I'm not so sure of it; a proof doesn't
come to mind.
|It is rather odd that the arguments typically
|given for Cantor's theorem are constructively acceptable, but the
|conclusion is not even understandable, intuitionistically (at least in
|the more interesting variants).
Huh? For every sequence of reals, there exists a real apart from
the terms of the sequence-- no problem.
|Even if you want to ignore such kinds of
|intuitionist mathematics as weird or irrelevant for 'real mathematics',
|doesn't this suggest that the classical argument sneaks in a number of
|hidden assumptions somewhere, and keeps unacceptably silent about it?
No, I don't think it even suggests it.
I remember long ago seeing a paper of Brouwer's in which he stated
his opinion on infinite cardinalities. If I remember correctly, he didn't
accept the existence of a power set of the continuum, but did accept
the continuum as a higher cardinality than the integers.
Keith Ramsay
Andrew Usher
This must be the result Russ Finlayson keeps trumpeting, claiming that
R is countable. One certainly could get that impression.
It shows to me that these models must be flawed, as there is only one
real R and only one real N, and Cantor's argument (either) that R is
uncountable doesn't rely upon ZFC. If any model adds more elements to
N, to make |N| = |R|, then we no longer have the natural numbers.
I don't understand how we can get more Cauchy sequences than we have.
If we redefined convergence, we'd no longer be talking about Cauchy
sequences.
Andrew Usher
Jesse F. Hughes
> je...@phiwumbda.org (Jesse F. Hughes) wrote in message news:<873c2ct...@phiwumbda.org>...
>> r...@tiki-lounge.com (Ross A. Finlayson) writes:
>>
>> >
>> > I have a theory where the powerset is the order type is the successor,
>> > for finite or infinite ordinals.
>>
>> ...
>>
>> What is the theory? What are the axioms?
>>
>> ...
>
> There are no axioms, that way it can be complete in extension of
> Goedel. It just uses zero/empty, tautology, and excluded middle,
> paraconsistently.
>
> That's what I've said for years.
No axioms. Good theory that.
Also "paraconsistent" logic. That's nice. That way, who cares if you
can prove a contradiction, eh? Makes your work a bit simpler.
What the heck is "zero/empty"? The existence of the empty set?
> You'll notice that N U {N} = N+1 = |N| = P(N) = succ(N), and that each
> powerset of N is not a limit ordinal, and that the only limit ordinal
> is N and 0.
Notice? I'll *notice* that? How does that follow from your empty
theory?
I don't think I can *notice* that.
The first equality is presumably definition. The second equality is,
strictly speaking, false in ZF[1] and certainly not provable in your
empty theory. The third equality (|N| = P(N)) is obviously false in
ZF and obviously not provable (and hence not "noticeable") in your
empty theory. The fourth equality is again false in ZF and clearly
not provable in your theory.
So, no, I *won't* notice that equation holds in your theory.
> You can read about the proper class consideration in "Ubiquitous
> Naturals, Infinity".
>
> Read!
Pass!
Sorry, Ross, wasting a little time on Usenet with you is already more
than I can justify.
Footnotes:
[1] Most people define |N| = N, so that |N+1| = |N| = N, but not
N+1 = |N|.
Herman Jurjus
> In article <2ogm7kF...@uni-berlin.de>, Herman Jurjus <h.ju...@hetnet.nl>
> writes:
> |Coming back to the original question: in (some forms of) intuitionist
> |mathematics, N has uncountable subsets, so 'larger size' has very little
> |to do with uncountability.
>
> I hope you're not confusing constructive mathematics with intuitionist
> mathematics.
Nope.
(But you're right, in intuitionism, it is a bit more controversial.)
> |It is rather odd that the arguments typically
> |given for Cantor's theorem are constructively acceptable, but the
> |conclusion is not even understandable, intuitionistically (at least in
> |the more interesting variants).
>
> Huh? For every sequence of reals, there exists a real apart from
> the terms of the sequence-- no problem.
Yes, but does that statement have any consequences for the 'size' of any
'set'? If yes, how and why. That part, imo, is missing in the typical
proofs.
--
Cheers,
Herman Jurjus
Herman Jurjus
[snip]
> For myself, the easiest way to understand generic extensions is through
> Boolean-valued models.
[rest snipped]
Thanks! So far i only knew (a little bit) about forcing.
I do have one question, though.
With this notion of extension, is it possible that an existing set in V
gets new elements in V[G]? Can you give me an example?
That is, suppose V is a (set-)model of ZFC and V[G] an extension as you
describe, and X is a set in V. So \hat X will be in V[G]. Can it occur
that for some x in V[G], x e \hat X, but for all y in V, x =/= \hat y?
I suppose it is intended that this is possible, but from the definition
you gave (of the embedding of V into V^B), i'm a little confused whether
this can occur or not.
> For x in V, we can define \hat x in V^B inductively by
>
> dom(\hat x) = {\hat y : y in x},
>
> \hat x(\hat y) = 1 for all y in x.
>
> This gives an embedding of V in V^B as a subclass.
>
> For x and y in V^B, ||x in y||, ||x subset y|| and ||x = y|| can be
> defined inductively by
>
> ||x in y|| = sup{ y(z).||x = z|| : z in dom(y) },
>
> ||x subset y|| = inf{ x(z)'+||z in y|| : z in dom(x) },
>
> ||x = y|| = ||x subset y||.||y subset x||,
>
> so that the "truth values" of "x in y", "x subset y" and "x = y" are given
> as elements of B.
>
> Specifically, it is demonstrable that
>
> ||\hat x in \hat y|| = 1 if x, y in V and x in y,
>
> ||\hat x in \hat y|| = 0 if x, y in V and x not in y,
[snip]
> For all x in V, i_G(\hat x) = x, so that V is a submodel of V[G].
[snip]
--
Cheers,
Herman Jurjus
Nath Rao
> It shows to me that these models must be flawed, as there is only one
> real R
Platonism and first-order logic are hard to reconcile. [This is
basically the point of S. Shapiro's book on second-order logic and
mathematics.]
To put it another way, to say that there is only one "real" R is to say
that there is only one universe of sets, so all these models are flawed
in that they must lack some property of the "real" V.
On the other hand, I think that Platonism is flawed, because it makes
categorical viewpoint hard to defend: If there is only one "real" V,
where does "the" category of sets live?
Let a proper class of V's bloom!
Nath Rao
Jesse F. Hughes
> Andrew Usher wrote:
>
>> It shows to me that these models must be flawed, as there is only one
>> real R
>
> Platonism and first-order logic are hard to reconcile.
That's a nice way of putting it.
--
But in our enthusiasm, we could not resist a radical overhaul of the
system, in which all of its major weaknesses have been exposed,
analyzed, and replaced with new weaknesses.
-- Bruce Leverett (presumably with apologies to Ambrose Bierce)
Andrew Usher
Your problem is that you keep trying to think logically about
Cantorian set theory, which inherently makes no sense. If you try to
make sense of it, you'll go mad.
Basically, infinite means goes on forever. When we say that N is
infinite, we mean that you'll never complete its enumeration. So
infinite sequences, sums, products, have this property.
This is how we define the cardinality of a set - we count it. Finite
sets are those for which we can finish the enumeration. Infinite sets,
then, may be defined as the union of an infinite sequence of larger
sets. When we say that R is uncountable, what we really mean is
'incomprehensible'. There's no rule giving you all of R.
Similarly an open interval is defined as the union of an infinite
sequence of intervals. So when we say [1,oo) we mean the union of
[1,a_n] where a is a sequence going to infinity. (0,1) is the union of
[a_n,b_n] where a goes to 0 and b goes to 1.
Be careful, though, that you actually can get such a sequence. For
example, Q is the smallest field in R. We can define infinitely many
other fields, depending on what other elements we admit. Now the union
of all such fields is S, the computable numbers. But S is not a set,
for there is no sequence of fields whose union gives S. In fact, S is
uncountable.
Andrew Usher
Andrew Usher
As long as you require functions to be piecewise continous (which is
more than reasonable), you don't need the continous nature of R. So
you could do analysis over Q, or any other countable dense set. It
still makes sense to pretend we're using the 'set' R, as it makes
things simpler.
Andrew Usher
Ron Sperber
Where did you get this definition of "set"? The real numbers in fact DO
form a set and can be well defined as equivalence classes of Cauchy
sequences of rationals. A set doesn't have to be countable at all.
In fact if you take a countably infinite set, it is easy to generate an
uncountable one by taking the power set of it.
-Ron
Jesse F. Hughes
> Perhaps I can try to correct you.
>
> Your problem is that you keep trying to think logically about
> Cantorian set theory, which inherently makes no sense. If you try to
> make sense of it, you'll go mad.
Gee, thanks. I get it now.
Wow, I must've been on the brink of madness there. Whew.
--
Jesse Hughes
"How come there's still apes running around loose and there are
humans? Why did some of them decide to evolve and some did not? Did
they choose to stay as a monkey or what?" -Kans. Board of Ed member
David McAnally
That R is uncountable is provable in ZF.
>If any model adds more elements to
>N, to make |N| = |R|,
Generic extensions don't add more elements to N. The ordinal omega is a
model for Peano's Axioms, with 0 being given by the empty set, and the
successor function being given by S(x) = x u {x}, where for sets A, B,
A u B denotes the union of A and B. Since any generic extension of a
model of ZFC has exactly the same ordinals as the original model, and
since omega is an ordinal, then it is impossible for any generic extension
to introduce new natural numbers, i.e. if N^V is N in the model V, and
N^{V[G]} is N in the generic extension V[G] of V, then N^{V[G]} = N^V.
|N| is not equal to |R| in any model of ZF.
What happens instead is that a generic extension introduces new functions,
so that, for example in this case I have already discussed, the generic
extension V[G] has bijections between the infinite sets A and B in V which
are not elements of V, in other words, the extension itself introduces new
bijections. The sets themselves are unaltered.
In the case of N and R, N passes through to the generic extension
unchanged (as noted above), so there are no new natural numbers, but a
generic extension can add new real numbers, so that while N and R^V may
be equinumerous in some generic extension V[G] of V, R^{V[G]} contains
some real numbers that R^V does not, and N and R^{V[G]} are not bijective
in V[G].
The Platonic attitude that there is exactly one set R of real numbers is
not fully appropriate. One cannot assume that there is exactly one set R
that must be common to all models of ZF, or that models of ZF are flawed
on the basis that R can be altered by taking a generic extension. Given
the Platonic genuine model of sets, and assuming that it is a model of ZF,
there are almost certainly subclasses of the genuine model which are
themselves models of ZF, in other words, we can refer to the elements of
a submodel as Sets (capital S), so that all Sets are sets, but not all
sets are Sets. All natural numbers are Sets, but it is possible that
some real numbers will be sets and not Sets. This means that the genuine
model contains as elements real numbers which the smaller subclass does
not.
A prime example of this is that many Platonists would consider that many
of the real numbers are not constructible (in the sense defined by Goedel
- the definition has nothing to do with straight edge and compass).
Since the subclass of constructible sets forms a model for ZFC, then there
is a model of ZFC such that R in the model is composed exclusively of
constructible real numbers. The model therefore has no nonconstructible
real numbers, but some generic extensions have nonconstructible real
numbers.
>then we no longer have the natural numbers.
>I don't understand how we can get more Cauchy sequences than we have.
It is because a Cauchy sequence is a function x from N to R such that for
all real d > 0, there exists k in N such that for all m, n in N such that
m > k and n > k, |x(m) - x(n)| < d. So whether or not a Cauchy sequence
is in a model V is dependent on whether the sequence itself (as a
function) is an element of V. The generic extension introduces new
functions into the model, and specifically, has the capability of
introducing new Cauchy sequences in Q. These new Cauchy sequences may
have limits that were not in the original model, and so may introduce
new real numbers.
>If we redefined convergence, we'd no longer be talking about Cauchy
>sequences.
We don't redefine convergence. Instead, the extension introduces new
functions that were not elements of the original model. Specifically,
it may introduce new sequences in Q (recalling that a sequence in Q is
a function from N to Q) that are not elements of the original model.
More specifically, it may introduce new Cauchy sequences in Q (which are
therefore functions from N to Q) which are not elements of the original
model.
David
-----
Herman Jurjus
[snip]
> Generic extensions don't add more elements to N.
Ok. I sort of suspected that.
[snip]
> What happens instead is that a generic extension introduces new functions,
> so that, for example in this case I have already discussed, the generic
> extension V[G] has bijections between the infinite sets A and B in V which
> are not elements of V, in other words, the extension itself introduces new
> bijections. The sets themselves are unaltered.
So, let's assume we have a set-based model of ZFC, in which N-in-V is
the 'real' N 'alfo' ('as looked at from the outside'), and, say, R-in-V
is uncountable-alfo. Then according to you there is an extension V' of V
in which N-in-V' is bijective to R-in-V'? (And judging from the
construction, V' is a set, at least a.l.f.o.)
But then i'd say the bijection-in-V' would also constitute a bijection
(a.l.f.o.), which is impossible, because that would be a bijection
between an uncountable set (R-in-V') and N (as N-in-V equals N-alfo).
So, i'd say your notion of extension can only make sense for, say,
countable models of ZFC, or at least with some restriction. (With
countable model, i mean countable a.l.f.o, of course.)
Unless... we have just proved that there exist no models of ZFC in which
N is the 'real' N and R is the 'real' R, a.l.f.o. That would be, well,
perhaps not shocking, but quite disturbing.
What am i missing, here?
--
Cheers,
Herman Jurjus
Ross A. Finlayson
> There are no axioms, that way it can be complete in extension of
> Goedel. It just uses zero/empty, tautology, and excluded middle,
> paraconsistently.
>
There are infinitely many and/or no axioms, in extension of Goedel and
myself.
> You'll notice that N U {N} = N+1 = |N| = P(N) = succ(N), and that each
> powerset of N is not a limit ordinal, and that the only limit ordinal
> is N and 0.
>
You'll notice that |N| should read "order type of N", and that the
only limit ordinal is 0, N, and Ord.
_______
/ \
| U |
\ /
^->0<-^
Diagram: All sets
Regards,
Ross F.
Jesse F. Hughes
>>
>> There are no axioms, that way it can be complete in extension of
>> Goedel. It just uses zero/empty, tautology, and excluded middle,
>> paraconsistently.
>>
>
> There are infinitely many and/or no axioms, in extension of Goedel and
> myself.
You're making no sense.
>
>> You'll notice that N U {N} = N+1 = |N| = P(N) = succ(N), and that each
>> powerset of N is not a limit ordinal, and that the only limit ordinal
>> is N and 0.
>>
>
> You'll notice that |N| should read "order type of N", and that the
> only limit ordinal is 0, N, and Ord.
And how should we "notice" that?
Is it derivable? Not from the theory you've given above.
Is it an axiom you forgot to mention? Well, since there are no other
axioms to speak of, that's not particularly interesting. Without some
constraint on the interpretation of the element-of relation, this
axiom doesn't say anything really.
Or is it one axiom among a number of axioms, contrary to what you've
said? Well, in this case, it *might* be interesting but we'd have to
see the other axioms. If this axiom allows a derivation of a
contradiction, however, then the resulting theory is again not very
interesting.
You don't get to say, "Oh, it's okay, I want a paraconsistent logic,"
as a justification for proposing a theory that leads to
contradiction. At least not unless you have an argument about why
paraconsistency is the right approach for a set theory.
You're just diddling, Ross. You have no theory here at all, from what
you've said. You're merely pretending that this flimsy intuition of
yours counts as a "theory".
Give us some axioms. Real axioms. In the form of formulas of a
formal language, preferably. Don't be vague about rules of deduction
either -- especially if you're going to rely on paraconsistency.
Either say *something* (of substance) or just shut up already.
--
Jesse F. Hughes
"That's what's annoying about Usenet as some loser will state a case,
get their ass kicked, but STILL keep coming back as if nothing
happened." -- James Harris explains his strategy.
Ross A. Finlayson
> r...@tiki-lounge.com (Ross A. Finlayson) writes:
>
> >>
> >> There are no axioms, that way it can be complete in extension of
> >> Goedel. It just uses zero/empty, tautology, and excluded middle,
> >> paraconsistently.
> >>
> >
> > There are infinitely many and/or no axioms, in extension of Goedel and
> > myself.
>
> You're making no sense.
>
meaningless." -- Jesse quoting me
Jesse, if set theory is to be the foundation, then a priori in set
theory there are only sets.
If a set is a container, then we have two points at which it can be
considered: the container of nothing and the container of everything.
One or the other or both is the only possible thing to consider, with
no other knowledge.
You know I've expressed preference for plain language statements of
logic, and that goes along with the concept that any explanation that
requires more than a single sentence or two is overcomplex, keep it
simple.
From zero, the empty set, we can consider the set containing it, and
the one containing that, and so on and so forth, ad infinitum.
Then as well we can consider the set of empty set and the set of empty
set, that's often called the ordinal one.
It is easy to continue with saying that every counting number is
represented as an ordinal by some construction of sets.
How is it that anothing non-empty exists in ZF? There is not an axiom
saying "exists empty", "exists set containing empty", "...". Now you
might demand an axiom to that effect, if so, then congratulations, you
require a new axiom. I don't.
Now in these ways of constructing ordinals from the empty set, we've
seen two simple methods, with 4 = {{{}}} + { {}, { {}, {} } }. Jesse,
2 + 2 = 4.
Another method to consider is a binary bitstring representing each
ordinal, as everything is an ordinal.
00000000...
Then, one set that is not necessarily the same as that is
11111111...
While that is so, in some meaning they represent the same value.
So anyways, I think that there are ways to have those equal, and as
well
11000000... = 2
and
01000000... = 2
In that way, I can rationally declare that in a theory of ubiquitous
ordinals, infinite sets are equivalent.
That's good for me, because then there are more reasons for EF to be a
bijective function between N and the unit interval of R, because it
is, and not not.
Now, here's a problem for you, in V[G], name an integer not in V.
Ross
Jesse F. Hughes
> You know I've expressed preference for plain language statements of
> logic, and that goes along with the concept that any explanation that
> requires more than a single sentence or two is overcomplex, keep it
> simple.
If you won't enumerate the axioms clearly and completely, we have
nothing at all to discuss.
You're diddling, pretending to think about deep issues when you don't
actually make explicit any of your fundamental assumptions.
> How is it that anothing non-empty exists in ZF? There is not an axiom
> saying "exists empty", "exists set containing empty", "...". Now you
> might demand an axiom to that effect, if so, then congratulations, you
> require a new axiom. I don't.
Because you're a moron, preferring unjustified and implicit assumption
rather than explicitly stating your axioms.
> Now, here's a problem for you, in V[G], name an integer not in V.
If I understood David's explanation, there ain't any.
--
Jesse Hughes
"[I]f gravel cannot make itself into an animal in a year, how could it
do it in a million years? The animal would be dead before it got
alive." --The Creation Evolution Encyclopedia
David McAnally
>r...@tiki-lounge.com (Ross A. Finlayson) writes:
<snip>
>Because you're a moron, preferring unjustified and implicit assumption
>rather than explicitly stating your axioms.
>> Now, here's a problem for you, in V[G], name an integer not in V.
>If I understood David's explanation, there ain't any.
That's correct. There are no new natural numbers in the extension, so
there are no new integers.
David
-----
David McAnally
>How is it that anothing non-empty exists in ZF? There is not an axiom
>saying "exists empty",
There is, actually. Or it is a theorem following from the Axiom of
Separation and the Axiom of Infinity (the latter being only invoked to
guarantee the existence of at least one set - the Axiom of Infinity can be
rewritten in such a manner that does not require the prior existence of an
empty set). The existence of an empty set also follows from the Axiom of
Infinity (which guarantees the existence of at least one set) and an
appropriate statement of the Axiom of Replacement.
>"exists set containing empty", "...".
Since ZF guarantees the existence of an empty set (which is unique in ZF
by the Axiom of Extensionality), the Axiom of the Power Set guarantees the
existence of a set which has the empty set as an element.
So the Axioms of ZF guarantee the existence opf an empty set and the
existence of sets which are not empty.
>Now you
>might demand an axiom to that effect, if so, then congratulations, you
>require a new axiom. I don't.
No new axiom required. ZF is already capable of guaranteeing the
existence of these sets.
<snip>
>Now, here's a problem for you, in V[G], name an integer not in V.
As Jesse pointed out, there are none. There are no new natural numbers,
David McAnally
>David McAnally wrote:
>[snip]
>> For myself, the easiest way to understand generic extensions is through
>> Boolean-valued models.
>[rest snipped]
>Thanks! So far i only knew (a little bit) about forcing.
>I do have one question, though.
>With this notion of extension, is it possible that an existing set in V
>gets new elements in V[G]? Can you give me an example?
No, an existing set in V does not get new elements in V[G]. For a set x
in V, since dom(\hat x) = {\hat y : y in V} and \hat x(\hat y) = 1 for
all y in V, then for any generic ultrafilter G,
i_G(\hat x) = {i_G(\hat y) : \hat y in dom(\hat x), \hat x(\hat y) in G}
= {i_G(\hat y) : y in x, 1 in G}
= {i_G(\hat y) : y in x}.
By e-induction, i_G(\hat x) = x for all x in V, and the elements of x are
unchanged in going from V to V[G]. If the cardinality of A is greater
than 1 and B is infinite, then the set of functions from A to B can
actually change between V and V[G], but the set of those functions which
are elements of V remains a set in V[G], and this is the set which is
identified with the original set of functions.
>That is, suppose V is a (set-)model of ZFC and V[G] an extension as you
>describe, and X is a set in V. So \hat X will be in V[G].
Actually, \hat X is in V^B, and its image in V[G] is i_G(\hat X), which is
equal to X. The elements of V^B are not sets in a model of ZFC, but their
images under i_G form a model of ZFC. Formulae and sentences are not
taken to be true or false in V^B, but are instead assigned an element of B
as the value. V[G] satisfies a formula iff the value of the formula in
V^B is an element of G.
>Can it occur
>that for some x in V[G], x e \hat X, but for all y in V, x =/= \hat y?
>I suppose it is intended that this is possible, but from the definition
>you gave (of the embedding of V into V^B), i'm a little confused whether
>this can occur or not.
I think that it is probably best to think of V^B as a skeleton on which to
hang a model of ZFC. It is not itself a model of ZFC (in the model-
theoretic sense), but the introduction of a generic ultrafilter gives it
the structure of a model. So V^B is an intermediary structure, built from
the model V and a Boolean algebra B in V (so that every element of V^B is
an element of V, all elements being functions). The choice of a generic
ultrafilter in B clothes the skeleton that is V^B and converts it into a
model V[G] of ZFC. This is done by mapping each element x of V^B to the
set of images of elements of dom(x) which x maps into an element of G,
i.e., for x in V^B, i_G(x) = {i_G(y) : y in dom(x), x(y) in G}. The model
is given by the class V[G] = {i_G(x) : x in V^B}.
The identification of \hat x in V^B corresponding x in V is just a means
of identifying an element of V^B which necessarily maps to x in V[G],
independently of the choice of G.
David
Ross A. Finlayson
bijection between N and R and thus P(N)? That's what it sounds like
to me.
Now we've seen in other threads that in IZF it is not inconsistent for
there to be bijections between N and R. Is that not so? Are there
thus consistent theories where the powerset mapping result is or is
not true, in a way as there are theories where the parallel postulate
is or is not true?
In a way that's similar to saying "there are functions between *N and
R, and by the way, *N is N, but don't tell people that because they
won't understand and it will thus anger and confuse them."
You can address the antidiagonal result and nested intervals result
thus showing that something like EF is necessary to map between the
naturals and the reals. The unit impulse function is a function.
About "empty exists", i.e., "nothing", "nothing at all", "absence of
anything", there might be theories with numbers before sets. There
also might be plain set theories with no further justification needed
than anything at all for that empty exists, and for the existence of
anything at all that so does something else, ad infinitum.
If you can adequately define hypercomplex numbers, you can completely
define all number theory.
What do you think of AC, aka Ono's Zorn's Lemma, the well-ordering
principle, etcetera. Is it implicit in CZF or IZF?
Warm regards, have a nice day, etc.,
Ross F.
David McAnally
>Herman Jurjus <h.ju...@hetnet.nl> writes:
>>David McAnally wrote:
>>[snip]
>>> For myself, the easiest way to understand generic extensions is through
>>> Boolean-valued models.
>>[rest snipped]
>>Thanks! So far i only knew (a little bit) about forcing.
>>I do have one question, though.
>>With this notion of extension, is it possible that an existing set in V
>>gets new elements in V[G]? Can you give me an example?
>No, an existing set in V does not get new elements in V[G]. For a set x
>in V, since dom(\hat x) = {\hat y : y in V} and \hat x(\hat y) = 1 for
>all y in V, then for any generic ultrafilter G,
Those last two lines should have read:
"in V, since dom(\hat x) = {\hat y : y in x} and \hat x(\hat y) = 1 for
all y in x, then for any generic ultrafilter G,"
>i_G(\hat x) = {i_G(\hat y) : \hat y in dom(\hat x), \hat x(\hat y) in G}
> = {i_G(\hat y) : y in x, 1 in G}
> = {i_G(\hat y) : y in x}.
David
-----
Andrew Usher
> > Generic extensions don't add more elements to N.
>
> Ok. I sort of suspected that.
So N is the same in all models ...
This is essentially the same as my beliefs. If a bijection can be
constructed between N in R[V] in some model V', then given that the
size of N, and of R[V], has not been changed, that would seem to say
that R[V] is countable, only unprovably so in V. Since we know that
the real R is not countable, that would seem to say that ZFC is
incomplete, or wrong.
Andrew Usher
Andrew Usher
> >> It shows to me that these models must be flawed, as there is only one
> >> real R
> >
> > Platonism and first-order logic are hard to reconcile.
>
> That's a nice way of putting it.
So we need second-order logic, right?
Andrew Usher
David McAnally
>So, you have, say, N in V and N in V[G],
And I stated that N in V and N in V[G] are equal to each other. There
are no natural numbers in V[G] which are not already natural numbers in V
(i.e. every natural number in V[G] is a natural number in V).
>and then you claim a
>bijection between N and R and thus P(N)?
No, I didn't. Supply a reference. I never claimed any bijection between
N and R, or between N and P(N). It is a theorem in ZF that there is no
bijection between N and P(N) (in fact, this is a corollary of the theorem
in ZF that there is no function which maps a set X onto P(X)), and since
it is a theorem of ZF that there is a bijection between R and P(N), then
it follows from ZF that there is no bijection between N and R.
>That's what it sounds like
>to me.
Then you got it wrong. Yes, I stated that if R^V is the set of real
numbers in a model V of ZFC, then there is a generic extension V[G]
in which there is a bijection between N and R^V. But R^V is not the set
of real numbers in V[G], but a proper subset of the set R^{V[G]} of real
numbers in V[G].
You seem to have some difficulty with some of the details of model theory.
The assertion of the existence of a bijection between N and R in a
specific model requires that the bijection itself be an element of the
model. Any extension of the model in which N and R in the model become
bijective also necessarily introduces sufficiently many new real numbers
that the natural numbers and real numbers are still not bijective in the
new model.
>Now we've seen in other threads that in IZF it is not inconsistent for
>there to be bijections between N and R. Is that not so?
No idea. I don't know about IZF.
>Are there
>thus consistent theories where the powerset mapping result is or is
>not true, in a way as there are theories where the parallel postulate
>is or is not true?
>In a way that's similar to saying "there are functions between *N and
>R, and by the way, *N is N, but don't tell people that because they
>won't understand and it will thus anger and confuse them."
The notation *N makes it look like you are referring to Robinson's
extension (or, as he refers to it in his book, 'enlargement') of N with
his introduction of nonstandard analysis. Yes, there are new elements of
*N which do not correspond to an element of N, but this is not the same
thing as a generic extension. Robinson's enlargement results from the
addition of extra axioms to the higher-order logic associated with the
first-order logic (or maybe additional axioms to the first-order logic,
itself), and the model is extended to accommodate the additional axioms.
In the case of a generic extension, the axioms remain the same, and the
model is extended to a new model which satisfies the same axioms.
>You can address the antidiagonal result and nested intervals result
>thus showing that something like EF is necessary to map between the
>naturals and the reals.
It is always possible to map from N to R. What you are wanting to discuss
is bijective mappings, of which there are none. Please provide a good
working definition of EF.
>The unit impulse function is a function.
What does the "unit impulse function" have to do with it?
>About "empty exists", i.e., "nothing", "nothing at all", "absence of
>anything", there might be theories with numbers before sets. There
>also might be plain set theories with no further justification needed
>than anything at all for that empty exists, and for the existence of
>anything at all that so does something else, ad infinitum.
What do you mean by "plain set theory"?
>If you can adequately define hypercomplex numbers, you can completely
>define all number theory.
Quaternions can be regarded as "hypercomplex".
>What do you think of AC, aka Ono's Zorn's Lemma, the well-ordering
>principle, etcetera. Is it implicit in CZF or IZF?
No idea about CZF or IZF, but not implicit in ZF, since Cohen demonstrated
the existence of models of ZF which do not satisfy the Axiom of Choice
(certain symmetric submodels of generic extensions of models of ZFC are
models of ZF but do not satisfy the Axiom of Choice).
David
-----
David McAnally
>David McAnally wrote:
>[snip]
>> Generic extensions don't add more elements to N.
>Ok. I sort of suspected that.
>[snip]
>> What happens instead is that a generic extension introduces new functions,
>> so that, for example in this case I have already discussed, the generic
>> extension V[G] has bijections between the infinite sets A and B in V which
>> are not elements of V, in other words, the extension itself introduces new
>> bijections. The sets themselves are unaltered.
>So, let's assume we have a set-based model of ZFC, in which N-in-V is
>the 'real' N 'alfo' ('as looked at from the outside'), and, say, R-in-V
>is uncountable-alfo. Then according to you there is an extension V' of V
>in which N-in-V' is bijective to R-in-V'? (And judging from the
>construction, V' is a set, at least a.l.f.o.)
As others have said, it is hard to reconcile Platonism and first-order
logic. As one with Platonist leanings, I have tried to work out what a
concrete version of the genuine objects would look like, without success.
>But then i'd say the bijection-in-V' would also constitute a bijection
>(a.l.f.o.), which is impossible, because that would be a bijection
>between an uncountable set (R-in-V') and N (as N-in-V equals N-alfo).
>So, i'd say your notion of extension can only make sense for, say,
>countable models of ZFC, or at least with some restriction. (With
>countable model, i mean countable a.l.f.o, of course.)
>Unless... we have just proved that there exist no models of ZFC in which
>N is the 'real' N and R is the 'real' R, a.l.f.o.
That might be the case.
>That would be, well,
>perhaps not shocking, but quite disturbing.
A fact which could be considered disturbing is that for any transfinite
cardinal not cofinal with omega, there exists a model of ZFC in which the
cardinality of R is the given cardinal. The upshot of this is that no
matter how many subsets of N you have in your model, more can always be
found to make larger models, and no matter how many real numbers you have
in a model, you can find an extension which has many more.
In fact, for any transfinite cardinal \lambda and any transfinite cardinal
\kappa whose cofinality exceeds \lambda, there is a model of ZFC in which
the power set of a set of cardinality \lambda has cardinality \kappa.
David
-----
David McAnally
>D.McAnally@i'm_a_gnu.uq.net.au (David McAnally) writes:
>>je...@phiwumbda.org (Jesse F. Hughes) writes:
>>>Herman Jurjus <h.ju...@hetnet.nl> writes:
<snip>
>>For myself, the easiest way to understand generic extensions is through
>>Boolean-valued models.
>An alternative approach uses forcing conditions. Let F be an ordered
>set in V. The elements of F are called conditions (or forcing
>conditions), and the condition q is said to be stronger than the condition
>p if q <= p (where q <= p denotes that q is less than or equal to p). Two
>conditions p and q are said to be compatible if there is a condition r
>which is stronger than both p and q, otherwise p and q are incompatible.
>F can be endowed with a topology, with an open base given by { [p] : p in
>F }, where [p] = { q in F : q <= p }. A subset A of F is open iff every
>condition stronger than an element of A is also an element of A. A subset
>A of F is closed iff every condition weaker than an element of A is also
>an element of A. Since a dense subset D of F is a subset with nonempty
>intersection with every nonempty open set, it follows that D is dense iff
>for all conditions p there exists a condition q stronger than p such that
>q is an element of D.
On a sidenote, if B is a Boolean algebra in V, let B-{0} be a set of
forcing conditions with the usual ordering, so that x is stronger than y
iff x <= y (x is less than or equal to y) in the usual Boolean ordering.
This means that x and y are compatible iff x.y is nonzero. A subset G
of B (G is not necessarily an element of V) is a generic ultrafilter in B
iff G is a generic set in B-{0}.
David
-----
Ross A. Finlayson
By "plain set theory" I meant theory where the only primary objects
were sets as opposed to a theory where numbers are primary objects.
So you have that in the generic extension that there are more elements
in R, but no more elements in N. R is the complete set of real
numbers in V, but you have that R in V is a proper subset of R in the
extension V[G]. Then you claim there is a bijection between N^V[G]
and R^[V]. N^V[G] has no more elements that N^V, yet you have it
bijectively mapping to R^V, which you claim has less elements than
R^V[G] as a proper subset or R^V. Conversely, R^V is the complete set
of real numbers, each real number is defined in R^V, and N^V[G] is not
different than N^V in that for each n E N^V there exists n E N^V[G]
and vice versa.
N^V and N^V[G] are the same set in that they differ in no elements,
they are each no different than N, yet you have that N^V[G]
bijectively maps to R^V but not to R^V[G], and claim that R^V[G] is a
proper superset of R^V. Please name or describe a real element of
R^V[G] not in R^V, that is R, the set of all real numbers.
That's why I interpreted what you said as affirming the existence of a
bijection between N and R, because N^V[G] has no element not in N and
bijectively maps to R, the set of all real numbers.
EF is basically n/oo defined for n E N. There is much discussion of
it on sci.math.
The average value of all hypercomplex numbers is zero.
I try to not use models in theory, because I think a theory should be
just a theory, and to not use classes in set theory, because a set
theory should have only sets.
Applying a metatheory or extended model or denying that anything is a
set in set theory only leads to obfuscation and not a real addressing
of the resolution of basically set existence issues within the set
theory itself. In that we attempt to address things infinite, merely
fabricating infinite ways to put off the resolution of the argument is
not a resolution. It is rather an expression, necessarily within the
presumed first-order theory, that serves the purpose of being a
cognitive reiteration of the presumed antinomous issue without
addressing it: the slippery slope, ever steeper.
To truly grasp or allow a resolution within a, and perhaps the only,
theory certainly allows the elucidation and enumeration of these
extensions as ornate embellishments of the theory, as examples of
logical capacity for abstraction within a theory sufficient to express
mathematical logic, but the theory offers a resolution, instead of
just ever more expansive non-explanations.
Regards,
Ross F.
Herman Jurjus
>>Unless... we have just proved that there exist no models of ZFC in which
>>N is the 'real' N and R is the 'real' R, a.l.f.o.
>
>
> That might be the case.
Sure.
But: using the very same argument, it would not be possible for a model
of ZFC to have an N that is countable-a.l.f.o, and an R-in-V that is
uncountable-a.l.f.o. And i don't think that is true (you actually
contradict it in the rest of your article, below).
I must be missing something, but what?
>>That would be, well,
>>perhaps not shocking, but quite disturbing.
>
>
> A fact which could be considered disturbing is that for any transfinite
> cardinal not cofinal with omega, there exists a model of ZFC in which the
> cardinality of R is the given cardinal.
Personally, i don't find it disturbing in itself that there are models
of ZFC in which the 'counterparts of real sets' are not the same as the
'real' Platonic ones. I got over that, long time ago. ;-)
But so far i thought of forcing as something that only worked for
countable models. You say your extension construction works in general,
and doesn't add elements to sets.
But then i see no alternative but to conclude that there exists no model
of ZFC in which N is countable-a.l.f.o and R is uncountable-a.l.f.o.
But as you show above, that must be wrong. What am i missing?
The upshot of this is that no
> matter how many subsets of N you have in your model, more can always be
> found to make larger models
But in that case, not every subset in the extension corresponds to a
subset in the outside world. For example in cases such as
V := {0} if CH and {1} if -CH
Fine, but that is a different thing, i think.
Which leads me to another question:
Does the fact that some object in any model V of ZFC is a bijection
between N-in-V and R-in-V (or two other V-sets, for that matter)
necessarily mean that that object represents a 'real' bijection between
the two V-sets, as looked at from the outside?
I think this should be true, as the definition of bijection is pretty
straightforward. But if it is true, my confusion still stands...
, and no matter how many real numbers you have
> in a model, you can find an extension which has many more.
>
> In fact, for any transfinite cardinal \lambda and any transfinite cardinal
> \kappa whose cofinality exceeds \lambda, there is a model of ZFC in which
> the power set of a set of cardinality \lambda has cardinality \kappa.
Yes, yes. But i'm not talking about powersets and their cardinalities.
We have a Tarski-structure, model of ZFC, with a countable N and an
uncountable R (both alfo). Now we extend the model without adding new
elements to each of these sets, so we end up with a model of ZFC, in
which there are two sets, still countable / uncountable alfo, but now
there is an object in the model of ZFC that is a bijection between the sets.
So, this object cannot be a bijection-a.l.f.o. (because the two sets are
not bijecive-a.l.f.o).
But: according to the Tarski definition of a model of ZFC, it will be a
bijection-a.l.f.o.
Because the definition of a bijection is of a simple form:
(for all.. there is... & if ..=.. then ..=..)
But, if that formula is true for some object in a model V of ZFC, that
object also constitutes a bijection between V-sets a.l.f.o. But that's
impossible, in this case.
Again, i must be missing something. But what?
--
Cheers,
Herman Jurjus
Herman Jurjus
>>N is the 'real' N and R is the 'real' R, a.l.f.o.
>
>
> That might be the case.
But i could just as well have said:
Unless... we have just proved that there exist no models of ZFC in which
N is uncountable a.l.f.o. (as looked from outside) and R is uncountable
a.l.f.o.
Perhaps that more clearly shows the disturbance.
>>That would be, well,
>>perhaps not shocking, but quite disturbing.
>
>
> A fact which could be considered disturbing is that for any transfinite
> cardinal not cofinal with omega, there exists a model of ZFC in which the
> cardinality of R is the given cardinal.
You mean that there are two cardinals in the model such that c1 = c2 in
the model? Or that R-in-V has the large cardinality, as looked at from
outside?
--
Cheers,
Herman Jurjus
KRamsay
r...@tiki-lounge.com (Ross A. Finlayson) writes:
|Now we've seen in other threads that in IZF it is not inconsistent for
|there to be bijections between N and R. Is that not so?
No, Cantor's theorem that for any function from N to R there exists
an element of R not in the image is a theorem of IZF. Please just
think of this as a mathematical fact, won't you?
I've seen proofs that it's consistent with various constructive formal
systems to assume that each function from N to N is computable.
If I'm not mistaken the same sort of proof works for IZF.
Assuming that each function from N to N is computable has a
number of consequences that are liable to be unfamiliar. Since it's
contradictory with the law of excluded middle, you have to be ready
to work with intuitionist logic.
Here, though, the relevant consequence is that the function mapping
those Turing machines that compute real numbers to the real numbers
they compute would be a SURJECTION from a SUBSET of the natural
numbers to the real numbers. Please, now that we've gone through this
more than once, either don't try to quote the result, or remember these
key details. It doesn't follow that there's a bijection. I don't happen to
know whether it's consistent to assume there exists a bijection between
R and a subset of N. I suspect not, but I didn't succeed in working it
out.
But most especially, there can't be a surjection from the whole of the
naturals to the reals.
Keith Ramsay
Herman Jurjus
But that is already well known for a long time.
There is no way to characterize 'the real model of ZFC' (whatever that
means) with new axioms or so. It was once known as the Skolem-Loewenheim
paradox.
But i think the issue i am touching is more serious.
If in some extension, there does exist a bijection between the sets,
and the sets have not received new elements, this necessarily means the
sets are 'really' bijective when looked upon from the outside world.
The only escape that i see, is that there is some restriction to David's
notion of generic extensions. (Like: the models have to be countable, or
transitive, or...)
(Unless ZFC is inconsistent, of course. :-) )
--
Cheers,
Herman Jurjus
Herman Jurjus
> A fact which could be considered disturbing is that for any transfinite
> cardinal not cofinal with omega, there exists a model of ZFC in which the
> cardinality of R is the given cardinal.
Sorry to bother you once more, but... can you provide a short argument
for this, as well?
Many thanks in advance,
--
Cheers,
Herman Jurjus
Herman Jurjus
> David McAnally wrote:
>
>> A fact which could be considered disturbing is that for any
>> transfinite cardinal not cofinal with omega, there exists a model of
>> ZFC in which the cardinality of R is the given cardinal.
>
>
> Sorry to bother you once more, but... can you provide a short argument
> for this, as well?
Moreover, what's the status of "there exists a transfinite cardinal not
cofinal with omega" ?
--
Cheers,
Herman Jurjus
Ross A. Finlayson
Cantor-Schroeder-Bernstein: it works both ways.
What that means is that one of the reasons that people call the reals
uncountable is because they've figured out a bijection between the
reals and the powerset of the naturals, thus they reason that there
are no bijections between the reals and the naturals, because
Cantor-Schroeder-Bernstein says the existence of a surjection either
way between two sets is proof of the existence of a bijection between
those two sets.
That is to say, the existence of a surjection from A to B and from B
to A implies that A and B are equivalent, and as well from A to B to C
and C to B to A through composition.
That implies it is not a mathematical fact and to promote the other
view as gospel, immutable, written in stone, etcetera, would thus be
deceitful. I'm rather angered that you would suggest the acceptance
of a mathematical falsehood as mathematical fact. Wouldn't you be?
Reexamine the claim about there being only one or none proper classes.
Consider why that demands dual representation of the ur-element as
both zero and Ord, regardless of whether Ord is N. (Steve, infinite
sets are equivalent.)
Unitize the analog! It's better to have a tool, even a primitive
tool, to measure sparse points of the continuum, than none, and bar
further consideration of the matter by fiat.
Why don't you consider that the direct sum of infinitely many copies
of N is zero?
From Dave Seaman's giveaway that an "uncountable" set is a countable
union of countable sets, to that the constructivist powerset mapping
argument is shoddy or fails, to these metatheoretical hand-wavings of
yes and no and ignorance fo the excluded middle, we've seen some of
what I would call progress in the state of discussion of these issues
here on sci.math, an open public forum where the participants are
often professional mathematicians and as such loathe to promote
untruth, particularly in the stark, concrete world of mathematics,
that is already a paradise with no need for the transfinite:
inconsistent and thus hellish.
So, in a post-Cantorian world, where it is accepted that Cantor's work
is among strong developments of foundations a hundred years ago,
Cantor is acknowledged as a discoverer and mathematician, and there's
only one way to biject the reals and naturals, and a horsedrawn
carriage leaves shit on the road, whereas an electric Fiat could go
300 miles per hour.
There are plainly rationalizations that you don't have to accept the
powerset mapping result of Cantor as mathematical fact. To do so is
more politically correct than honest for the urbane sophisticate
uncaring of the truth, or card-carrying party member, except that it
is as well pretentious and vapid, and the seasons change.
You're intelligent and understand the basic tenets of rational
discourse, and by now you're probably familiar with the rather few
arguments of transfinite set theory: assume to argue that infinite
sets are equivalent.
You may well be able to have it both ways, in a post-Cantorian world.
You have a choice.
The numbers are uncaring. Only software cares.
Have a nice day. Regards,
Ross A. Finlayson