ZFC Can Prove Everything Provable? But how do you define "provable" here??
Charlie-Boo
What is so nice about the statement "ZFC (whatevuh) can prove
everything provable." is:
1. It shows we (prof.s) have everything figured out - that can be
figured out, that is. Hilbert said, "We can prove everything that is
true in a consistent system." and was wrong thanks to Godel.
(Actually, Cantor laid the groundwork but nobody - including Hilbert -
applied it to WFFs and Sets of Numbers until Godel) So now we have a
new way to say it: "We can prove everything that can be proven."
2. It's so vague that you can't disprove (refute) it.
Since I like to take things that are designed to be too complex to
refute and then show a simple way to refute it (e.g. a horribly
complex explanation of a Program Synthesis system: "Where's the PHP
program to list the prime numbers?" - see Einstein "dear skeptic"):
3. How do you define "provable" here?
C-B
A start: We can prove 2<3 since x<y is an r.e. relation i.e. there is
a wff (we can call it "<" if we want) w1 such that ( x < y ) iff |-w1
(x,y) so |-2<3 since 2<3. (Isn't that a much better proof than the
constructive one that gives the proof?) But (2,3) is in lots of
sets . . .
Jesse F. Hughes
> What is so nice about the statement "ZFC (whatevuh) can prove
> everything provable." is: [...]
You act as if this is a claim that people have made. I've never heard
anyone make such a claim and it seems trivially false on the plainest
interpretation. (ZFC does not prove the theorems of non-well-founded
set theory, for instance.)
Can you find a single mathematical text that has claimed this? Or
any other source that makes this claim?
--
Jesse F. Hughes
"C is for Cookie. That's good enough for me."
Cookie Monster
Victor Porton
> Charlie-Boo <shymath...@gmail.com> writes:
> > What is so nice about the statement "ZFC (whatevuh) can prove
> > everything provable." is: [...]
>
> You act as if this is a claim that people have made. I've never heard
> anyone make such a claim and it seems trivially false on the plainest
> interpretation. (ZFC does not prove the theorems of non-well-founded
> set theory, for instance.)
But can't the theorems of non-well-founded set theory be mapped into
ZFC theorems? (I don't know and even doubt whether I can formulate
this question exactly.)
Jesse F. Hughes
> On Jan 5, 8:44 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> Charlie-Boo <shymath...@gmail.com> writes:
>> > What is so nice about the statement "ZFC (whatevuh) can prove
>> > everything provable." is: [...]
>>
>> You act as if this is a claim that people have made. I've never heard
>> anyone make such a claim and it seems trivially false on the plainest
>> interpretation. (ZFC does not prove the theorems of non-well-founded
>> set theory, for instance.)
>
> But can't the theorems of non-well-founded set theory be mapped into
> ZFC theorems? (I don't know and even doubt whether I can formulate
> this question exactly.)
As I understand it, one can translate the formulas of ZFA into
formulas of ZFC so that the translated axioms of ZFA are theorems of
ZFC.
I don't see how this supports the claim that ZFC proves everything
that is provable. What we have here, of course, is an everyday straw
man argument. Charlie pretends that this claim is common and then
proceeds to argue that it is too vague to be meaningful. But this is
obvious and uninteresting, since I've never seen *anyone* claim that
ZFC proves everything that is provable.
>
>> Can you find a single mathematical text that has claimed this? Or
>> any other source that makes this claim?
So, Charlie, how about it?
--
"This page contains information of a type (text/html) that can only be
viewed with the appropriate Plug-in. Click OK to download Plugin."
--- Netscape 4.7 error message
David C. Ullrich
<je...@phiwumbda.org> wrote:
>Charlie-Boo <shyma...@gmail.com> writes:
>
>> What is so nice about the statement "ZFC (whatevuh) can prove
>> everything provable." is: [...]
>
>You act as if this is a claim that people have made. I've never heard
>anyone make such a claim and it seems trivially false on the plainest
>interpretation.
Well of course it's false. He refuted it!
Some people are never satisfied... CB says utterly wrong things
and people complain. He finally says something true, and not
only that it's something obviously true, and people complain
about that too.
Giggle.
porky_...@my-deja.com
> On Tue, 05 Jan 2010 13:44:36 -0500, "Jesse F. Hughes"
>
> <je...@phiwumbda.org> wrote:
>
> >> What is so nice about the statement "ZFC (whatevuh) can prove
> >> everything provable." is: [...]
>
> >You act as if this is a claim that people have made. I've never heard
> >anyone make such a claim and it seems trivially false on the plainest
> >interpretation.
>
> Well of course it's false. He refuted it!
>
> Some people are never satisfied... CB says utterly wrong things
> and people complain. He finally says something true, and not
> only that it's something obviously true, and people complain
> about that too.
>
> Giggle.
>
>
>
Even the broken clock shows the correct time twice a day. shall we
rejoice every time it happens?
David C. Ullrich
whoosh...
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)
Robert
Charlie-Boo
Make that "in Ordinary Mathematics".
http://groups.google.com/group/sci.logic/msg/3b4c60e5742d025f?hl=en
C-B
Jesse F. Hughes
No, even Lord Boetian didn't explicitly say that. Instead, he tried
to get your claim somewhat closer to the truth.
In any case, Lord Boetian isn't really the sort of citation I was
looking for. Is there any case in which this claim has been published
(you know, where editors are involved)? Or any case in which a
professional mathematician (preferably with some recognition in the
field) has said this?
--
"Now for once I might actually have an audience that realizes that
[my proof of Fermat's Last Theorem is correct], because you see,
they'll finally know what's in it for them--cold, hard cash."
--James Harris embarks on a new mathematical strategy.
Charlie-Boo
> > What is so nice about the statement "ZFC (whatevuh) can prove
> > everything provable." is: [...]
>
> You act as if this is a claim that people have made. I've never heard
> anyone make such a claim and it seems trivially false on the plainest
> interpretation. (ZFC does not prove the theorems of non-well-founded
> set theory, for instance.)
>
> Can you find a single mathematical text that has claimed this? Or
> any other source that makes this claim?
“The development of mathematics towards greater exactness has, as is
well-known, lead
to formalization of large areas of it such that you can carry out
proofs by following a few
mechanical rules. The most comprehensive current formal systems are
the system of
Principia Mathematica (PM) on the one hand, the Zermelo-Fraenkelian
axiom-system
of set theory on the other hand. These two systems are so far
developed that you
can formalize in them all proof methods that are currently in use in
mathematics.”
On formally undecidable propositions of Principia Mathematica and
related systems I - Kurt Godel, 1931
C-B
Andrew Usher
> Charlie-Boo <shymath...@gmail.com> writes:
> > What is so nice about the statement "ZFC (whatevuh) can prove
> > everything provable." is: [...]
>
> You act as if this is a claim that people have made. I've never heard
> anyone make such a claim and it seems trivially false on the plainest
> interpretation. (ZFC does not prove the theorems of non-well-founded
> set theory, for instance.)
>
> Can you find a single mathematical text that has claimed this? Or
> any other source that makes this claim?
He gave a passage from Goedel that could be interpreted so. But, he
(Goedel) proved that no first order theory can 'prove everything
provable'.
Andrew Usher
Jesse F. Hughes
> �The development of mathematics towards greater exactness has, as is
> well-known, lead
> to formalization of large areas of it such that you can carry out
> proofs by following a few
> mechanical rules. The most comprehensive current formal systems are
> the system of
> Principia Mathematica (PM) on the one hand, the Zermelo-Fraenkelian
> axiom-system
> of set theory on the other hand. These two systems are so far
> developed that you
> can formalize in them all proof methods that are currently in use in
> mathematics.�
>
> On formally undecidable propositions of Principia Mathematica and
> related systems I - Kurt Godel, 1931
Yes, all proof methods can be formalized in those systems.
That does *not* mean that they prove everything provable. What he
means is that all of the usual bits of informal mathematical reasoning
that was being used in 1931 could be made formal in these two formal
systems.
--
Jesse F. Hughes
"Have we learned nothing, nothing, from the downfall of Vanilla Ice?"
-- Time Magazine columnist Lev Grossman on
James Frey's /A Million Pieces/.
Charlie-Boo
> On Jan 5, 12:44 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>
> > Charlie-Boo <shymath...@gmail.com> writes:
> > > What is so nice about the statement "ZFC (whatevuh) can prove
> > > everything provable." is: [...]
>
> > You act as if this is a claim that people have made. I've never heard
> > anyone make such a claim and it seems trivially false on the plainest
> > interpretation. (ZFC does not prove the theorems of non-well-founded
> > set theory, for instance.)
>
> > Can you find a single mathematical text that has claimed this? Or
> > any other source that makes this claim?
>
> He gave a passage from Goedel that could be interpreted so.
What is your alternate interpretation? (BTW It is the first sentences
of Godel's famous article - not famous enough, I guess.)
> But, he
> (Goedel) proved that no first order theory can 'prove everything
> provable'.
Then you can answer my question: How do you define "everything
provable"?
C-B
> Andrew Usher
scattered
> Charlie-Boo <shymath...@gmail.com> writes:
> > What is so nice about the statement "ZFC (whatevuh) can prove
> > everything provable." is: [...]
>
> You act as if this is a claim that people have made. I've never heard
> anyone make such a claim and it seems trivially false on the plainest
> interpretation. (ZFC does not prove the theorems of non-well-founded
> set theory, for instance.)
>
> Can you find a single mathematical text that has claimed this? Or
> any other source that makes this claim?
> --
It seems to me that Charlie-Boo *is* alluding to a fairly wide spread
albeit informal view. Namely, that mathematics is ultimately reducible
to set theory and that ZFC captures what can be nonproblematically
proved about sets. To study the limits of provability in ZFC is
(roughly) equivalent to studying the limits of provability in
mathematics. Why else would the study of ZFC be so central in
metamathematics? (cf. Wikidpedia's first sentence in its entry on ZFC:
"Zermelo–Fraenkel set theory with the axiom of choice, commonly
abbreviated ZFC, is the standard form of axiomatic set theory and as
such is the most common foundation of mathematics" - what is meant by
a foundation?) Most mathematicians take the independence of CH from
ZFC as implying that the truth or falsity of CH will likely forever
remain a matter of mathematical speculation rather than mathematical
knowledge. Things like category theory shows that the situation is
more complicated than C.B. allows, but his claim is nevertheless
approximately true rather than "trivially false." This doesn't imply
that the rest of his post makes much sense.
-scattered
Andrew Usher
> > But, he
> > (Goedel) proved that no first order theory can 'prove everything
> > provable'.
>
> Then you can answer my question: How do you define "everything
> provable"?
I would say that it must be taken in an informal sense, to mean
everything that we can write a proof for. For example, we can prove
that the real numbers are larger than any countable set, but
Loewenheim-Skolem says we can't do it in ZFC (or any first order
theory).
Andrew Usher
Andrew Usher
> Yes, all proof methods can be formalized in those systems.
>
> That does *not* mean that they prove everything provable. What he
> means is that all of the usual bits of informal mathematical reasoning
> that was being used in 1931 could be made formal in these two formal
> systems.
How are these two statements not contradictory? If all proofs can be
formalised in ZFC, then every proof is a proof in ZFC, etc.
Andrew Usher
Rotwang
> [...]
>
> Make that "in Ordinary Mathematics".
>
> http://groups.google.com/group/sci.logic/msg/3b4c60e5742d025f?hl=en
Funny thread. I particularly like the bit where you ask where in
Goldrei's /Classic Set Theory/ you can find a ZFC proof of something
outside arithmetic and set theory. Presumably you failed to notice that
the whole second chapter is devoted to the construction of the real
numbers (including a proof of the completeness axiom), or that the ZF
axioms are used to retrospectively justify the steps in Chapter 2
shortly after they are introduced in Chapter 4 (e.g. Exercise 4.35 in
the first edition).
Jesse F. Hughes
I suppose it depends on what you mean. Let's take ZFA, for instance,
the theory of non-well-founded sets. Admittedly, Goedel was *not*
talking about this theory (since no one was talking about ZFA in
1931), but let's see in what sense "you can formalize in them all
proof methods" that are used in ZFA.
You can do so in this way, as I recall: interpret the sets of ZFA as
particular kinds of graphs. Graphs can easily be represented in ZFC.
This re-interpretation induces a translation of the language of ZFA
into the language of ZFC (where the epsilon relation of ZFA is *not*
the epsilon relation of ZFC). Under this interpretation, the axioms
of ZFA are mapped to theorems of ZFC. Since the underlying logic
(namely FOL=) is the same for both theories, it follows that every
theorem of ZFA is mapped to a theorem of ZFC.
In this sense, the reasoning of ZFA can be formalized in ZFC. Really,
this is not so different than our usual interpretation of PA in ZFC.
This is what I think that Goedel had in mind. I don't see any
contradiction here, nor do I think that this formalization is
adequately captured by saying that "ZFC proves everything that is
provable in ordinary mathematics." It seems to me that this latter
statement is very misleading.
--
Jesse F. Hughes
"To be honest, I don't have enough interest in math to spend the time
it would take to clean up the mess that I believe has been created in
the past 100 or so years." -- Curt Welch lets the world down.
Jesse F. Hughes
Loewenheim-Skolem says *what* now?
You're plainly mistaken. ZFC *does* prove that R is uncountable.
--
Jesse F. Hughes
"If the world weren't rather strange, by now I should at least be with
some research group talking about my number theory research."
-- James S. Harris learns the world is a funny place
Jesse F. Hughes
>> But, he
>> (Goedel) proved that no first order theory can 'prove everything
>> provable'.
>
> Then you can answer my question: How do you define "everything
> provable"?
Why should anyone define this term which has been used by no one but
you?
--
"At the Microsoft-sponsored cocktail reception in the Galaxy Ballroom
that evening, Robert Dees urges us 'to network on behalf of the people
of Iraq,'"
-- Naomi Klein reports on Microsoft's efforts to further democracy.
Jesse F. Hughes
> On Jan 5, 1:44 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> Charlie-Boo <shymath...@gmail.com> writes:
>> > What is so nice about the statement "ZFC (whatevuh) can prove
>> > everything provable." is: [...]
>>
>> You act as if this is a claim that people have made. I've never heard
>> anyone make such a claim and it seems trivially false on the plainest
>> interpretation. (ZFC does not prove the theorems of non-well-founded
>> set theory, for instance.)
>>
>> Can you find a single mathematical text that has claimed this? Or
>> any other source that makes this claim?
>> --
>
> It seems to me that Charlie-Boo *is* alluding to a fairly wide spread
> albeit informal view. Namely, that mathematics is ultimately reducible
> to set theory and that ZFC captures what can be nonproblematically
> proved about sets.
Yes, certainly he's alluding to this view, but his attempt to express
this view was vague nonsense. As a result, he criticized a statement
that no one has made -- in other words, we have a simple straw man
argument. His own quote from Goedel was rather less problematic than
a claim that "ZFC proves everything provable" (even though the quote
itself is fairly vague and Goedel took for granted that his audience
understood his meaning).
In particular, neither Goedel nor anyone else I've read said anything
at all about "everything provable", yet it is this phrase that Charlie
is focusing on.
> To study the limits of provability in ZFC is (roughly) equivalent to
> studying the limits of provability in mathematics. Why else would
> the study of ZFC be so central in metamathematics? (cf. Wikidpedia's
> first sentence in its entry on ZFC: "Zermelo-Fraenkel set theory
> with the axiom of choice, commonly abbreviated ZFC, is the standard
> form of axiomatic set theory and as such is the most common
> foundation of mathematics" - what is meant by a foundation?) Most
> mathematicians take the independence of CH from ZFC as implying that
> the truth or falsity of CH will likely forever remain a matter of
> mathematical speculation rather than mathematical knowledge. Things
> like category theory shows that the situation is more complicated
> than C.B. allows, but his claim is nevertheless approximately true
> rather than "trivially false." This doesn't imply that the rest of
> his post makes much sense.
No, his claim as stated is trivially false in the plainest
interpretation. If he meant a more subtle interpretation, then
perhaps he should explain.
--
Jesse F. Hughes
"If mathematics doesn't recognize its social dependencies, then
perpetual slavery is just around the corner."
-- Han de Bruijn, on why set theory is a capitalist tool
David C. Ullrich
No, LS does not say that.
>Andrew Usher
Andrew Usher
> > I would say that it must be taken in an informal sense, to mean
> > everything that we can write a proof for. For example, we can prove
> > that the real numbers are larger than any countable set, but
> > Loewenheim-Skolem says we can't do it in ZFC (or any first order
> > theory).
>
> Loewenheim-Skolem says *what* now?
>
> You're plainly mistaken. ZFC *does* prove that R is uncountable.
You keep saying this but it's plainly false in real logic. If ZFC has
a model where 'R' is countable - and LS says it must - then ZFC can
not establish its uncountability as that would make a contradiction.
Sure, we can _write_ a proof in ZFC but it can't be sound.
Andrew Usher
Andrew Usher
> >For example, we can prove
> >that the real numbers are larger than any countable set, but
> >Loewenheim-Skolem says we can't do it in ZFC (or any first order
> >theory).
>
> No, LS does not say that.
See my response to Jesse Hughes. LS does say that every first-order
theory has a countable model, which implies my statement.
Andrew Usher
Andrew Usher
> > How are these two statements not contradictory? If all proofs can be
> > formalised in ZFC, then every proof is a proof in ZFC, etc.
>
> I suppose it depends on what you mean. Let's take ZFA, for instance,
> the theory of non-well-founded sets. Admittedly, Goedel was *not*
> talking about this theory (since no one was talking about ZFA in
> 1931), but let's see in what sense "you can formalize in them all
> proof methods" that are used in ZFA.
>
> You can do so in this way, as I recall: interpret the sets of ZFA as
> particular kinds of graphs. Graphs can easily be represented in ZFC.
> This re-interpretation induces a translation of the language of ZFA
> into the language of ZFC (where the epsilon relation of ZFA is *not*
> the epsilon relation of ZFC). Under this interpretation, the axioms
> of ZFA are mapped to theorems of ZFC. Since the underlying logic
> (namely FOL=) is the same for both theories, it follows that every
> theorem of ZFA is mapped to a theorem of ZFC.
>
> In this sense, the reasoning of ZFA can be formalized in ZFC. Really,
> this is not so different than our usual interpretation of PA in ZFC.
Then, if that's correct, every proof in ZFA is a proof in ZFC as well.
> This is what I think that Goedel had in mind. I don't see any
> contradiction here, nor do I think that this formalization is
> adequately captured by saying that "ZFC proves everything that is
> provable in ordinary mathematics." It seems to me that this latter
> statement is very misleading.
How, exactly? Isn't that why set theory was invented?
Andrew Usher
Jesse F. Hughes
> On Jan 8, 8:31 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>
>> > How are these two statements not contradictory? If all proofs can be
>> > formalised in ZFC, then every proof is a proof in ZFC, etc.
>>
>> I suppose it depends on what you mean. Let's take ZFA, for instance,
>> the theory of non-well-founded sets. Admittedly, Goedel was *not*
>> talking about this theory (since no one was talking about ZFA in
>> 1931), but let's see in what sense "you can formalize in them all
>> proof methods" that are used in ZFA.
>>
>> You can do so in this way, as I recall: interpret the sets of ZFA as
>> particular kinds of graphs. Graphs can easily be represented in ZFC.
>> This re-interpretation induces a translation of the language of ZFA
>> into the language of ZFC (where the epsilon relation of ZFA is *not*
>> the epsilon relation of ZFC). Under this interpretation, the axioms
>> of ZFA are mapped to theorems of ZFC. Since the underlying logic
>> (namely FOL=) is the same for both theories, it follows that every
>> theorem of ZFA is mapped to a theorem of ZFC.
>>
>> In this sense, the reasoning of ZFA can be formalized in ZFC. Really,
>> this is not so different than our usual interpretation of PA in ZFC.
>
> Then, if that's correct, every proof in ZFA is a proof in ZFC as
> well.
No, each proof in ZFA corresponds to a proof in ZFC through a
particular translation of the language of ZFA to the language of ZFC.
>
>> This is what I think that Goedel had in mind. I don't see any
>> contradiction here, nor do I think that this formalization is
>> adequately captured by saying that "ZFC proves everything that is
>> provable in ordinary mathematics." It seems to me that this latter
>> statement is very misleading.
>
> How, exactly? Isn't that why set theory was invented?
It's misleading because it neglects the actual situation: the proofs
depend on translating one formal system into another.
Jesse F. Hughes
Wrong. The proof is clearly sound and it's conclusion true.
The statement "R is not countable" is true iff there is no surjection
N -> R. In models of ZFC, there is no surjection (in the model) from
the interpretation of N to the interpretation of R.
In any case, what the hell would you mean that "we can write a proof
in ZFC but it can't be sound"? The only way in which a deductive
argument is unsound is if it is invalid or one of its premises false.
Surely you agree that the argument R is uncountable is valid, so which
premise do you doubt? (We may as well leave out choice, since it
isn't used in this proof. Similar for foundation/regularity.)
--
"Flowers in the Attic" was based on a true story. [...] HOW is it OK
to just butcher such an awesome piece of work? It's like passing
Pokemon off as the Mona-Lisa; sick and entirely wrong.
-- An Amazon reviewer pissed that the movie didn't include incest
David C. Ullrich
<k_over...@yahoo.com> wrote:
>On Jan 8, 2:42 pm, David C. Ullrich <ullr...@math.okstate.edu> wrote:
>
>> >For example, we can prove
>> >that the real numbers are larger than any countable set, but
>> >Loewenheim-Skolem says we can't do it in ZFC (or any first order
>> >theory).
>>
>> No, LS does not say that.
>
>See my response to Jesse Hughes. LS does say that every first-order
>theory has a countable model,
Yes it does. (Well, every first-order theory in a countable
language, which includes ZFC.)
> which implies my statement.
No it doesn't. "the real numbers are larger than any countable set"
is a theorem of ZFC.
And hence, in any model of ZFC it is true that there is no function
mapping N onto R. Now. Say M is a countable model of ZFC.
It's true that there is function _in M_ mapping N onto R
(or rather mapping what M thinks is N onto what M thinks
is R). The fact that M itself is countable does not contradict
this - M is countable, so there is a mapping from the natural
numbers onto what M thinks is R. But that mapping is not
an element of M.
>Andrew Usher
Andrew Usher
> >> In this sense, the reasoning of ZFA can be formalized in ZFC. Really,
> >> this is not so different than our usual interpretation of PA in ZFC.
>
> > Then, if that's correct, every proof in ZFA is a proof in ZFC as
> > well.
>
> No, each proof in ZFA corresponds to a proof in ZFC through a
> particular translation of the language of ZFA to the language of ZFC.
But that's enough to establish that ZFA is consistent (if ZFC is).
> >> This is what I think that Goedel had in mind. I don't see any
> >> contradiction here, nor do I think that this formalization is
> >> adequately captured by saying that "ZFC proves everything that is
> >> provable in ordinary mathematics." It seems to me that this latter
> >> statement is very misleading.
>
> > How, exactly? Isn't that why set theory was invented?
>
> It's misleading because it neglects the actual situation: the proofs
> depend on translating one formal system into another.
Well I think then we only disagree semantically.
Andrew Usher
Andrew Usher
> > You keep saying this but it's plainly false in real logic. If ZFC has
> > a model where 'R' is countable - and LS says it must - then ZFC can
> > not establish its uncountability as that would make a contradiction.
> > Sure, we can _write_ a proof in ZFC but it can't be sound.
>
> Wrong. The proof is clearly sound and it's conclusion true.
>
> The statement "R is not countable" is true iff there is no surjection
> N -> R. In models of ZFC, there is no surjection (in the model) from
> the interpretation of N to the interpretation of R.
Yes, but the surjection (in countable models) still _exists_, we just
can't find it in ZFC. The conclusion that *N < *R is then false.
> In any case, what the hell would you mean that "we can write a proof
> in ZFC but it can't be sound"? The only way in which a deductive
> argument is unsound is if it is invalid or one of its premises false.
> Surely you agree that the argument R is uncountable is valid, so which
> premise do you doubt?
In logic, when you prove that A -> B, exhibiting just one possible
world where A is true and B false contradicts the proof. Countable
models do the same thing for ZFC's proof, even though the conclusion
is true: the real real numbers are not countable.
Andrew Usher
Andrew Usher
> "The real numbers are larger than any countable set"
> is a theorem of ZFC.
>
> And hence, in any model of ZFC it is true that there is no function
> mapping N onto R. Now. Say M is a countable model of ZFC.
> It's true that there is function _in M_ mapping N onto R
I suppose you meant a function not in M.
> (or rather mapping what M thinks is N onto what M thinks
> is R). The fact that M itself is countable does not contradict
> this - M is countable, so there is a mapping from the natural
> numbers onto what M thinks is R. But that mapping is not
> an element of M.
This just shows the incompleteness of ZFC if such situations can
exist.
Andrew Usher
Jesse F. Hughes
> On Jan 8, 6:20 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>
>> >> In this sense, the reasoning of ZFA can be formalized in ZFC. Really,
>> >> this is not so different than our usual interpretation of PA in ZFC.
>>
>> > Then, if that's correct, every proof in ZFA is a proof in ZFC as
>> > well.
>>
>> No, each proof in ZFA corresponds to a proof in ZFC through a
>> particular translation of the language of ZFA to the language of ZFC.
>
> But that's enough to establish that ZFA is consistent (if ZFC is).
Er, yes, of course.
>> >> This is what I think that Goedel had in mind. I don't see any
>> >> contradiction here, nor do I think that this formalization is
>> >> adequately captured by saying that "ZFC proves everything that is
>> >> provable in ordinary mathematics." It seems to me that this latter
>> >> statement is very misleading.
>>
>> > How, exactly? Isn't that why set theory was invented?
>>
>> It's misleading because it neglects the actual situation: the proofs
>> depend on translating one formal system into another.
>
> Well I think then we only disagree semantically.
Yes, we were arguing over whether a particular expression adequately
captures the situation. Hence, it is a semantic disagreement.
--
Jesse F. Hughes
"With [President Bush] endorsing [Intelligent Design], at the very
least it makes Americans who have that position more respectable, for
lack of a better phrase." -- Gary L. Bauer, in search of a thesaurus
Jesse F. Hughes
>> (or rather mapping what M thinks is N onto what M thinks
>> is R). The fact that M itself is countable does not contradict
>> this - M is countable, so there is a mapping from the natural
>> numbers onto what M thinks is R. But that mapping is not
>> an element of M.
>
> This just shows the incompleteness of ZFC if such situations can
> exist.
Nothing at all to do with incompleteness.
--
Jesse F. Hughes
"You see 300 of something, anything, and you go `[Man], that's a lot of
stuff.'" -- Jim Bigler, quoted in the Pittsburgh Post-Gazette.
Jesse F. Hughes
> On Jan 9, 6:21 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>
>> > You keep saying this but it's plainly false in real logic. If ZFC has
>> > a model where 'R' is countable - and LS says it must - then ZFC can
>> > not establish its uncountability as that would make a contradiction.
>> > Sure, we can _write_ a proof in ZFC but it can't be sound.
>>
>> Wrong. The proof is clearly sound and it's conclusion true.
>>
>> The statement "R is not countable" is true iff there is no surjection
>> N -> R. In models of ZFC, there is no surjection (in the model) from
>> the interpretation of N to the interpretation of R.
>
> Yes, but the surjection (in countable models) still _exists_, we just
> can't find it in ZFC. The conclusion that *N < *R is then false.
No.
The proposition |N| < |R| is true iff there is no surjection in the
model from the interpretation of N onto the interpretation of R.
Hence, this proposition is true in every model.
>> In any case, what the hell would you mean that "we can write a proof
>> in ZFC but it can't be sound"? The only way in which a deductive
>> argument is unsound is if it is invalid or one of its premises false.
>> Surely you agree that the argument R is uncountable is valid, so which
>> premise do you doubt?
>
> In logic, when you prove that A -> B, exhibiting just one possible
> world where A is true and B false contradicts the proof. Countable
> models do the same thing for ZFC's proof, even though the conclusion
> is true: the real real numbers are not countable.
You didn't answer my question. An argument is unsound only if either
it is invalid or at least one of its premises is false. So, which is
it?
(In any case, the fact is that |N| < |R| is true in every model of ZFC
in the requisite sense, namely, that in each model, there is no
surjection [[N]] -> [[R]].)
--
Jesse F. Hughes
"Most people don't even know what a rootkit is, so why should they
care about it."
-- Thomas Hesse, sony executive defends DRM-by-rootkit.
David C. Ullrich
<k_over...@yahoo.com> wrote:
>On Jan 9, 9:04�am, David C. Ullrich <ullr...@math.okstate.edu> wrote:
>
>> "The real numbers are larger than any countable set"
>> is a theorem of ZFC.
>>
>> And hence, in any model of ZFC it is true that there is no function
>> mapping N onto R. Now. Say M is a countable model of ZFC.
>> It's true that there is function _in M_ mapping N onto R
>
>I suppose you meant a function not in M.
What I actually intended in that sentence was "there is no function
in M...". Yes, it was certainly a typo.
>> (or rather mapping what M thinks is N onto what M thinks
>> is R). The fact that M itself is countable does not contradict
>> this - M is countable, so there is a mapping from the natural
>> numbers onto what M thinks is R. But that mapping is not
>> an element of M.
>
>This just shows the incompleteness of ZFC if such situations can
>exist.
ZFC us certainly incomplete. But that has nothing to do with
what we're talking about here - if T is a complete extension
of ZFC (which of course cannot be recursively axiomatizable)
then everything we've said applies equally well to T, even
though T is complete.
>Andrew Usher
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)
Andrew Usher
> >> The statement "R is not countable" is true iff there is no surjection
> >> N -> R. In models of ZFC, there is no surjection (in the model) from
> >> the interpretation of N to the interpretation of R.
>
> > Yes, but the surjection (in countable models) still _exists_, we just
> > can't find it in ZFC. The conclusion that *N < *R is then false.
>
> No.
>
> The proposition |N| < |R| is true iff there is no surjection in the
> model from the interpretation of N onto the interpretation of R.
That statement is absurd and offensive to logic. The notion of one-to-
one correspondence, and therefore equinumerosity, precedes any formal
theory.
> Hence, this proposition is true in every model.
True in ZFC but not True philosophically.
> >> In any case, what the hell would you mean that "we can write a proof
> >> in ZFC but it can't be sound"? The only way in which a deductive
> >> argument is unsound is if it is invalid or one of its premises false.
> >> Surely you agree that the argument R is uncountable is valid, so which
> >> premise do you doubt?
>
> > In logic, when you prove that A -> B, exhibiting just one possible
> > world where A is true and B false contradicts the proof. Countable
> > models do the same thing for ZFC's proof, even though the conclusion
> > is true: the real real numbers are not countable.
>
> You didn't answer my question. An argument is unsound only if either
> it is invalid or at least one of its premises is false. So, which is
> it?
The (implied) premiss that is false is that N and R in ZFC are the
real N and R. If you say that we should not use this premiss, then you
are admitting that ZFC can say nothing about the real N and R, and
that therefore the proof 'R is larger than any countable set' can not
be formalised in ZFC - which was my original claim.
Andrew Usher
Daryl McCullough
>> The proposition |N| < |R| is true iff there is no surjection in the
>> model from the interpretation of N onto the interpretation of R.
>
>That statement is absurd and offensive to logic. The notion of one-to-
>one correspondence, and therefore equinumerosity, precedes any formal
>theory.
How does that make what Jesse said absurd? I really don't understand
exactly what your complaint is. The claim |N| < |R| is true, and it's
provable in ZFC. The point of ZFC was to have a theory capable of proving
true facts about sets. So what's the problem?
>> Hence, this proposition is true in every model.
>
>True in ZFC but not True philosophically.
What's not true? The claim that |N| < |R|? It certainly is true.
>The (implied) premiss that is false is that N and R in ZFC are the
>real N and R.
That's not a premise of ZFC. ZFC is a *theory*, a collection of statements.
It doesn't *have* any sets, it *describes* sets. It doesn't make sense
to talk about "N and R in ZFC" because N and R are not elements of ZFC,
they are described by ZFC. And for most things, the descriptions provided
by ZFC are accurate.
Now, the existence of nonstandard models (such as countable models) shows
that ZFC's descriptions don't uniquely pin down R. That's an interesting
fact, but it certainly isn't relevant to the question of soundness of ZFC.
Soundness is about whether the statements ZFC proves about N or R are in
fact true.
>If you say that we should not use this premiss, then you
>are admitting that ZFC can say nothing about the real N and R, and
>that therefore the proof 'R is larger than any countable set' can not
>be formalised in ZFC - which was my original claim.
ZFC certainly does say something about the real N and R, namely that
|N| < |R|. You can write down a proof of that statement in the language
of ZFC. What more do you want from a theory of sets?
--
Daryl McCullough
Ithaca, NY
Jesse F. Hughes
> On Jan 11, 6:58 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>
>> >> The statement "R is not countable" is true iff there is no surjection
>> >> N -> R. In models of ZFC, there is no surjection (in the model) from
>> >> the interpretation of N to the interpretation of R.
>>
>> > Yes, but the surjection (in countable models) still _exists_, we just
>> > can't find it in ZFC. The conclusion that *N < *R is then false.
>>
>> No.
>>
>> The proposition |N| < |R| is true iff there is no surjection in the
>> model from the interpretation of N onto the interpretation of R.
>
> That statement is absurd and offensive to logic. The notion of one-to-
> one correspondence, and therefore equinumerosity, precedes any formal
> theory.
Er, right. So, modern mathematical semantics generally is absurd and
offensive to logic. After all, the interpretation of the proposition
|N| < |R| follows directly from the definition of model.
>> Hence, this proposition is true in every model.
>
> True in ZFC but not True philosophically.
Whatever that means. The fact is that mathematical theories are
interpreted in structures and inherit their meaning via their
interpretations. That's simply how things are.
>
>> >> In any case, what the hell would you mean that "we can write a proof
>> >> in ZFC but it can't be sound"? The only way in which a deductive
>> >> argument is unsound is if it is invalid or one of its premises false.
>> >> Surely you agree that the argument R is uncountable is valid, so which
>> >> premise do you doubt?
>>
>> > In logic, when you prove that A -> B, exhibiting just one possible
>> > world where A is true and B false contradicts the proof. Countable
>> > models do the same thing for ZFC's proof, even though the conclusion
>> > is true: the real real numbers are not countable.
>>
>> You didn't answer my question. An argument is unsound only if either
>> it is invalid or at least one of its premises is false. So, which is
>> it?
>
> The (implied) premiss that is false is that N and R in ZFC are the
> real N and R. If you say that we should not use this premiss, then you
> are admitting that ZFC can say nothing about the real N and R, and
> that therefore the proof 'R is larger than any countable set' can not
> be formalised in ZFC - which was my original claim.
No, this isn't the conclusion. Let's suppose there is a real (i.e.,
intended) model for ZFC, in which [[N]] is the "real" N and [[R]] is
the "real" R. Then the proof |N| < |R| shows that there is no
surjection from the "real" N onto the "real" R -- which is just what
you wanted to know.
As it happens, it also shows that there is no surjection from N to R
in a countable model of ZFC, but so what?
In any case, your "implied" premise is nonsense. It is not part of
the formal argument, and it is the formal argument you called
unsound. So, try again.
--
"This is based on the assumption that the difference in set size is what
makes the important difference between finite and infinite sets, but I think
most people -- even the mathematicians -- will agree that that probably
isn't the case." -- Allan C Cybulskie explains infinite sets
Andrew Usher
> No, this isn't the conclusion. Let's suppose there is a real (i.e.,
> intended) model for ZFC, in which [[N]] is the "real" N and [[R]] is
> the "real" R.
OK, I'll accept that. But ...
> Then the proof |N| < |R| shows that there is no
> surjection from the "real" N onto the "real" R -- which is just what
> you wanted to know.
This is wrong, I think. The proof only shows that there is no
surjection within ZFC; I don't see why we have to believe that the
standard model includes all possible mappings, given that non-standard
models don't. The reason we know that N < R is the informal or non-
firstorderisable argument; a formal argument within set theory can
only be less certain.
Andrew Usher
Jesse F. Hughes
> On Jan 12, 9:08 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>
>> No, this isn't the conclusion. Let's suppose there is a real (i.e.,
>> intended) model for ZFC, in which [[N]] is the "real" N and [[R]] is
>> the "real" R.
>
> OK, I'll accept that. But ...
>
>> Then the proof |N| < |R| shows that there is no
>> surjection from the "real" N onto the "real" R -- which is just what
>> you wanted to know.
>
> This is wrong, I think. The proof only shows that there is no
> surjection within ZFC; I don't see why we have to believe that the
> standard model includes all possible mappings, given that non-standard
> models don't.
The so-called standard model is supposed to include all the sets that
"really" exist. If so, then it includes all functions which "really"
exist and hence there is no surjection from the "real" N onto the
"real" R.
> The reason we know that N < R is the informal or non-
> firstorderisable argument; a formal argument within set theory can
> only be less certain.
I don't know what you mean by "the non-firstorderisable argument"
(though I'm mighty fond of the adjective).
In any model of ZFC, there is no surjection from [[N]] onto [[R]].
If we suppose that there is a universe containing every set, it is
surely a model of ZFC and hence it follows that in this universe (the
"real" universe of sets), there is no surjection from [[N]] (the
"real" N) onto [[R]] (the "real" R).
And that's what you wanted to know.
Your complaint, it seems, is not that ZFC fails to prove what you
want, but that it also proves other claims which don't interest you.
Namely, we see that there is no surjection [[N]] -> [[R]] in any model
of ZFC. Since the universe of sets is a model of ZFC, the fact you're
interested in follows. But you seem to be bugged by the observation
that this fact also follows in non-standard models, even though there
is nothing at all controversial about its truth in non-standard
models, once one realizes that the claim is relativized to the
functions existing in such models.
--
Jesse F. Hughes
"I'm not good at math(s) - but even I know zero is nothing. To suggest
zero isn't nothing is madness."
-- Phantom scojocup, on why 0 * 4 = 4
Math1723
> What is so nice about the statement "ZFC (whatevuh) can prove
> everything provable." is:
First of all, ZFC cannot "prove everything provable". For example,
ZFC cannot prove some things provable from ZFC+CH, or some things from
ZFC+~CH. ZFC, however, can prove all those things provable from ZFC.
(We know this from the Reflexive Law of Equality.)
(Sheesh, Charlie, sometimes you just need to think about what you
wrote, man.)
Andrew Usher
> > This is wrong, I think. The proof only shows that there is no
> > surjection within ZFC; I don't see why we have to believe that the
> > standard model includes all possible mappings, given that non-standard
> > models don't.
>
> The so-called standard model is supposed to include all the sets that
> "really" exist.
Is this provable? If not, how is it not just blind faith?
> If so, then it includes all functions which "really"
> exist and hence there is no surjection from the "real" N onto the
> "real" R.
> > The reason we know that N < R is the informal or non-
> > firstorderisable argument; a formal argument within set theory can
> > only be less certain.
>
> I don't know what you mean by "the non-firstorderisable argument"
> (though I'm mighty fond of the adjective).
I didn't invent it. Anyway, what I mean is that the 'real' N and R
can't be defined by first-order logic (an immediate consequence of
LS), so the informal proof (e.g. Cantor's argument) that N < R can't
be, either. What I mean is that seeing it formalised, or knowing it
can be formalised, in ZFC does not make one any more certain that R is
not countable; on the other hand, it is open to objections (such as
mine, above) while the informal argument is not.
So what is set theory really good for, anyway?
Andrew Usher
Jesse F. Hughes
> On Jan 12, 10:36 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>
>> > This is wrong, I think. The proof only shows that there is no
>> > surjection within ZFC; I don't see why we have to believe that the
>> > standard model includes all possible mappings, given that non-standard
>> > models don't.
>>
>> The so-called standard model is supposed to include all the sets that
>> "really" exist.
>
> Is this provable? If not, how is it not just blind faith?
That's what the standard model *is*. It's the universe of all sets.
(Well, there may be minor issues regarding the definition of model
here, since what I call "the standard model" is class-sized, but let
us ignore these technicalities.)
>> If so, then it includes all functions which "really"
>> exist and hence there is no surjection from the "real" N onto the
>> "real" R.
>
>> > The reason we know that N < R is the informal or non-
>> > firstorderisable argument; a formal argument within set theory can
>> > only be less certain.
>>
>> I don't know what you mean by "the non-firstorderisable argument"
>> (though I'm mighty fond of the adjective).
>
> I didn't invent it. Anyway, what I mean is that the 'real' N and R
> can't be defined by first-order logic (an immediate consequence of
> LS), so the informal proof (e.g. Cantor's argument) that N < R can't
> be, either. What I mean is that seeing it formalised, or knowing it
> can be formalised, in ZFC does not make one any more certain that R is
> not countable; on the other hand, it is open to objections (such as
> mine, above) while the informal argument is not.
I don't know the details, but it seems to me that if there is a
universe of sets, then it is a model of ZFC and in that model of ZFC,
the interpretation of N is the "real" set of natural numbers (and R
the "real" set of reals). The interpretation of N and R in ZFC
involves quantification over all the objects in its model and hence
over all sets.
But I'll defer to someone who knows more about ZFC than I do for this
argument.
> So what is set theory really good for, anyway?
If you're not interested in studying the use and limits of formal
arguments, not so much.
--
"Britney thought the idea of a pre-nup was vile, because she is
loved-up with Kevin and cannot envisage breaking up. However, [...] no
one in Hollywood these days get married without brokering a
deal. [...] She had a long chat with Kevin and he was cool about it."
H. J. Sander Bruggink
>> The so-called standard model is supposed to include all the sets that
>> "really" exist.
>
> Is this provable? If not, how is it not just blind faith?
You have to start somewhere, yes. You have to make some basic
assumptions about the "real" sets. These basic assumptions are called
axioms. The whole point about using a formal system is that it is
completely clear what the axioms are. If the axioms of ZFC are actually
true of sets, then the model of "real" sets is indeed a model of ZFC.
[snip]
> I didn't invent it. Anyway, what I mean is that the 'real' N and R
> can't be defined by first-order logic (an immediate consequence of
> LS), so the informal proof (e.g. Cantor's argument) that N< R can't
> be, either. What I mean is that seeing it formalised, or knowing it
> can be formalised, in ZFC does not make one any more certain that R is
> not countable; on the other hand, it is open to objections (such as
> mine, above) while the informal argument is not.
The informal argument needs axioms also. It just doesn't spell them out.
So you can still ask: is it provable that the informal argument is about
"real sets"? If not, how is it not just blind faith?
groente
-- Sander
Andrew Usher
wrote:
> You have to start somewhere, yes. You have to make some basic
> assumptions about the "real" sets. These basic assumptions are called
> axioms. The whole point about using a formal system is that it is
> completely clear what the axioms are. If the axioms of ZFC are actually
> true of sets, then the model of "real" sets is indeed a model of ZFC.
This is circular. Defining the 'real sets' in terms of ZFC makes no
sense because, clearly, the 'real sets' do exist independently of any
formal theory we may write.
> > I didn't invent it. Anyway, what I mean is that the 'real' N and R
> > can't be defined by first-order logic (an immediate consequence of
> > LS), so the informal proof (e.g. Cantor's argument) that N< R can't
> > be, either. What I mean is that seeing it formalised, or knowing it
> > can be formalised, in ZFC does not make one any more certain that R is
> > not countable; on the other hand, it is open to objections (such as
> > mine, above) while the informal argument is not.
>
> The informal argument needs axioms also. It just doesn't spell them out.
> So you can still ask: is it provable that the informal argument is about
> "real sets"?
Those axioms, though, are simply definition. We all agree, even when
not defining them formally, that N is the unique structure satisfying
Peano's axioms (or equivalent) and that R is the unique complete
Archimedean ordered field. Those definitions are not provable because
they are definition. But the formalisation in set theory requires
those _and_ that ZFC correctly models them - a strictly stronger
assumption.
Andrew Usher
Andrew Usher
> >> The so-called standard model is supposed to include all the sets that
> >> "really" exist.
>
> > Is this provable? If not, how is it not just blind faith?
>
> That's what the standard model *is*. It's the universe of all sets.
> (Well, there may be minor issues regarding the definition of model
> here, since what I call "the standard model" is class-sized, but let
> us ignore these technicalities.)
So, in other words, it's just the largest possible model of ZFC?
> I don't know the details, but it seems to me that if there is a
> universe of sets, then it is a model of ZFC and in that model of ZFC,
> the interpretation of N is the "real" set of natural numbers (and R
> the "real" set of reals). The interpretation of N and R in ZFC
> involves quantification over all the objects in its model and hence
> over all sets.
Granting that there is a model of ZFC with the 'real' N and R, though,
how do we know that that same model also includes all possible
functions from N to R? And of course we can't 'quantify over all
sets'.
Andrew Usher
David C. Ullrich
<k_over...@yahoo.com> wrote:
>On Jan 14, 8:41�am, "H. J. Sander Bruggink" <brugg...@uni-due.de>
>wrote:
>
>> You have to start somewhere, yes. You have to make some basic
>> assumptions about the "real" sets. These basic assumptions are called
>> axioms. The whole point about using a formal system is that it is
>> completely clear what the axioms are. If the axioms of ZFC are actually
>> true of sets, then the model of "real" sets is indeed a model of ZFC.
>
>This is circular. Defining the 'real sets' in terms of ZFC makes no
>sense because, clearly, the 'real sets' do exist independently of any
>formal theory we may write.
How in the world can you read the previous paragraph as a supposed
_definition_ of the "real sets"?
Daryl McCullough
>Granting that there is a model of ZFC with the 'real' N and R, though,
>how do we know that that same model also includes all possible
>functions from N to R? And of course we can't 'quantify over all
>sets'.
Actually, we know this because there is a definable 1-1 function
between elements of R and elements of N -> R. So any model of ZFC will
have this function, which means that if it has all elements of R, it
will have all elements of N -> R, as well.
Daryl McCullough
>
>On Jan 14, 8:41=A0am, "H. J. Sander Bruggink" <brugg...@uni-due.de>
>wrote:
>> The informal argument needs axioms also. It just doesn't spell them out.
>> So you can still ask: is it provable that the informal argument is about
>> "real sets"?
>
>Those axioms, though, are simply definition. We all agree, even when
>not defining them formally, that N is the unique structure satisfying
>Peano's axioms (or equivalent) and that R is the unique complete
>Archimedean ordered field. Those definitions are not provable because
>they are definition. But the formalisation in set theory requires
>those _and_ that ZFC correctly models them - a strictly stronger
>assumption.
Sure, ZFC is stronger than the theory of naturals, or the theory of
reals. That's *why* people use ZFC, because it doesn't just talk about
natural, or just talk about reals, but can talk about sets of reals,
and sets of sets of reals, and sets of sets of sets of reals, etc.
Jesse F. Hughes
> On Jan 14, 8:36 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>
>> >> The so-called standard model is supposed to include all the sets that
>> >> "really" exist.
>>
>> > Is this provable? If not, how is it not just blind faith?
>>
>> That's what the standard model *is*. It's the universe of all sets.
>> (Well, there may be minor issues regarding the definition of model
>> here, since what I call "the standard model" is class-sized, but let
>> us ignore these technicalities.)
>
> So, in other words, it's just the largest possible model of ZFC?
I don't know whether there are larger possible models of ZFC. My only
(informal) claim is that the universe of all sets is a model of ZFC.
>> I don't know the details, but it seems to me that if there is a
>> universe of sets, then it is a model of ZFC and in that model of ZFC,
>> the interpretation of N is the "real" set of natural numbers (and R
>> the "real" set of reals). The interpretation of N and R in ZFC
>> involves quantification over all the objects in its model and hence
>> over all sets.
>
> Granting that there is a model of ZFC with the 'real' N and R, though,
> how do we know that that same model also includes all possible
> functions from N to R? And of course we can't 'quantify over all
> sets'.
We are not merely granting that some model has sets (isomorphic to)
the real N and R, but rather that
(1) The universe of all sets contains sets (isomorphic to?) the real N
and R and
(2) The universe of all sets is a model of ZFC.
Now, any function from N to R is a subset of N x R and hence an
element of the universe of all sets. Thus, this same model has all
possible functions N -> R.
(Again, we are ignoring the technical difficulty that models are
usually defined as having a carrier set, while here we are speaking of
a class-sized carrier. In any case, the set of all functions N -> R
exists in certain set-sized models of ZFC, if I'm not mistaken.)
As far as your last remark (we can't 'quantify over all sets'), I'm
mighty confused. ZFC does just that. Perhaps you have never seen the
axioms of ZFC?
--
"Now I'm informing all of you that the people arguing against me are EVIL,
yes they are real, live EVIL people as mathematics is that important, so
it's important enough for Evil itself to send minions like them."
-- James Harris on Evil's interest in Algebraic Number Theory
Andrew Usher
> > So, in other words, it's just the largest possible model of ZFC?
>
> I don't know whether there are larger possible models of ZFC. My only
> (informal) claim is that the universe of all sets is a model of ZFC.
And I suppose such a larger model, which you claim might exist,
includes sets not in the universe of all sets??
> > Granting that there is a model of ZFC with the 'real' N and R, though,
> > how do we know that that same model also includes all possible
> > functions from N to R? And of course we can't 'quantify over all
> > sets'.
>
> We are not merely granting that some model has sets (isomorphic to)
> the real N and R, but rather that
>
> (1) The universe of all sets contains sets (isomorphic to?) the real N
> and R and
>
> (2) The universe of all sets is a model of ZFC.
These axioms assume that we have some definition of 'the universe of
all sets' independent of ZFC.
> Now, any function from N to R is a subset of N x R and hence an
> element of the universe of all sets. Thus, this same model has all
> possible functions N -> R.
But wouldn't this argument apply to all models, not only the standard
one. In that case, countable models could not exist (because N < R
would then be false in them), and you have acknowledged they do.
> As far as your last remark (we can't 'quantify over all sets'), I'm
> mighty confused. ZFC does just that. Perhaps you have never seen the
> axioms of ZFC?
I meant, we can't do it in a model-independent way.
Andrew Usher
Andrew Usher
wrote:
> >Those axioms, though, are simply definition. We all agree, even when
> >not defining them formally, that N is the unique structure satisfying
> >Peano's axioms (or equivalent) and that R is the unique complete
> >Archimedean ordered field. Those definitions are not provable because
> >they are definition. But the formalisation in set theory requires
> >those _and_ that ZFC correctly models them - a strictly stronger
> >assumption.
>
> Sure, ZFC is stronger than the theory of naturals, or the theory of
> reals.
No. Unlike the informal arguments, it is not categorical.
Also, a theory that requires more assumptions is logically _weaker_.
Andrew Usher
Andrew Usher
> >This is circular. Defining the 'real sets' in terms of ZFC makes no
> >sense because, clearly, the 'real sets' do exist independently of any
> >formal theory we may write.
>
> How in the world can you read the previous paragraph as a supposed
> _definition_ of the "real sets"?
It's the best we can do, given the inherent logical concept. I take
Goedel's theorem as meaning that no formal system can characterise
everything that's real, because it would have to be consistent and
complete to do that.
Andrew Usher
Jesse F. Hughes
Are you sure you read David's question? Because I don't see how this
was any sort of an answer.
--
"Sexual love makes of the loved person an Object of appetite; as soon
as that appetite has been stilled, the person is cast aside as one
casts away a lemon which has been sucked dry." -- Immanuel Kant
"Squeeze my lemon til the juice runs down my leg." -- Robert Johnson
Jesse F. Hughes
> On Jan 15, 10:53 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>
>> > So, in other words, it's just the largest possible model of ZFC?
>>
>> I don't know whether there are larger possible models of ZFC. My only
>> (informal) claim is that the universe of all sets is a model of ZFC.
>
> And I suppose such a larger model, which you claim might exist,
> includes sets not in the universe of all sets??
I made no claim that a larger model exists. Of course, if a larger
model exists, then it would include objects which are not sets (in the
same way that non-standard models of PA include objects which are not
natural numbers).
"I don't know whether X is true," does not mean "X might be true".
>> > Granting that there is a model of ZFC with the 'real' N and R, though,
>> > how do we know that that same model also includes all possible
>> > functions from N to R? And of course we can't 'quantify over all
>> > sets'.
>>
>> We are not merely granting that some model has sets (isomorphic to)
>> the real N and R, but rather that
>>
>> (1) The universe of all sets contains sets (isomorphic to?) the real N
>> and R and
>>
>> (2) The universe of all sets is a model of ZFC.
>
> These axioms assume that we have some definition of 'the universe of
> all sets' independent of ZFC.
I don't really want to make any claims regarding Platonism/Realism.
My comments on the universe of all sets is really just a matter of
explaining my argument in your terms.
For my own part, what I know about sets comes from the axioms of ZFC.
One consequence of those axioms is that, in any model of ZFC, there is
no surjection N -> R. That is all I mean when I say that R is
uncountable.
Admittedly, some models of ZFC are countable. That's pretty odd, no
doubt. But it certainly does not make the claim "R is uncountable"
suspect, since R *is* uncountable in the requisite sense in *every*
model.
You, on the other hand, are motivated by the view that there is a
"real" set R and a "real" set N (and, one supposes, a real universe of
sets containing R and N). Let us suppose so. I reckon that this
universe is a model of ZFC -- do you disagree? From these
suppositions, it follows that there is no surjection N -> R in the
real universe of sets either.
Do you see anything at all controversial about these claims from your
perspective?
>> Now, any function from N to R is a subset of N x R and hence an
>> element of the universe of all sets. Thus, this same model has all
>> possible functions N -> R.
>
> But wouldn't this argument apply to all models, not only the standard
> one. In that case, countable models could not exist (because N < R
> would then be false in them), and you have acknowledged they do.
No, because countable models do not include every function N -> R.
>> As far as your last remark (we can't 'quantify over all sets'), I'm
>> mighty confused. ZFC does just that. Perhaps you have never seen the
>> axioms of ZFC?
>
> I meant, we can't do it in a model-independent way.
I'm not sure what that means, but if you agree that there is a
universe of sets and this universe is a model of ZFC, then the formal
proof that R is uncountable shows that there is no surjection N -> R
in the universe of sets.
Isn't that what you were interested in?
--
Jesse F. Hughes
"Radicals are interesting because they were considered 'radical' by
the people who played with them who wrote a lot of math work that
modern mathematics depends on." --Another JSH history lesson
Jesse F. Hughes
> Also, a theory that requires more assumptions is logically _weaker_.
If you mean "axioms" where you write "assumptions", then, no, this is
not how the terms "stronger" and "weaker" are used in logic.
In the simplest case, where two theories T and T' share a common
language, then T is stronger[1] than T' if, whenever T' |- P, then
T |- P.
Footnotes:
[1] "at least as strong as", if you prefer, since this is a weak
ordering.
--
Jesse F. Hughes
"Well, you know as soon as you have a new number I will be happy to
add it to the list. Don't try those childish tit-for-tat games with
me." -- Ross Finlayson on Cantor's theorem.
Daryl McCullough
>
>On Jan 15, 8:05 am, stevendaryl3...@yahoo.com (Daryl McCullough)
>wrote:
>
>> >Those axioms, though, are simply definition. We all agree, even when
>> >not defining them formally, that N is the unique structure satisfying
>> >Peano's axioms (or equivalent) and that R is the unique complete
>> >Archimedean ordered field. Those definitions are not provable because
>> >they are definition. But the formalisation in set theory requires
>> >those _and_ that ZFC correctly models them - a strictly stronger
>> >assumption.
>>
>> Sure, ZFC is stronger than the theory of naturals, or the theory of
>> reals.
>
>No. Unlike the informal arguments, it is not categorical.
What does that mean?
Look, I have no idea what you are going on about, so I guess I
should just drop out of the conversation. The point of having
a foundational theory such as ZFC is that it provides a unified
framework that allows us to capture the informal arguments from
a wide variety of fields. ZFC works very well in this respect;
it is generally believed that the types of mathematical objects
occurring in almost all of mathematics (number theory, analysis,
function theory, probability theory, differential equations, etc.)
can be formulated in set theoretic language such that the standard
theorems of those fields become theorems of ZFC.
I understand that the Lowenheim Skolem theorem is causing you
angst about the whole endeavor, but I really don't understand
why.
>Also, a theory that requires more assumptions is logically _weaker_.
I'm using "stronger theory" in the sense of "proves more theorems".
David Bernier
> On Jan 15, 10:53 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>
>>> So, in other words, it's just the largest possible model of ZFC?
>> I don't know whether there are larger possible models of ZFC. My only
>> (informal) claim is that the universe of all sets is a model of ZFC.
>
> And I suppose such a larger model, which you claim might exist,
> includes sets not in the universe of all sets??
>
>>> Granting that there is a model of ZFC with the 'real' N and R, though,
>>> how do we know that that same model also includes all possible
>>> functions from N to R? And of course we can't 'quantify over all
>>> sets'.
>> We are not merely granting that some model has sets (isomorphic to)
>> the real N and R, but rather that
>>
>> (1) The universe of all sets contains sets (isomorphic to?) the real N
>> and R and
>>
>> (2) The universe of all sets is a model of ZFC.
>
> These axioms assume that we have some definition of 'the universe of
> all sets' independent of ZFC.
No. Some experience with axiomatic systems in maths is enough to show
the fundamental distinction between the "foundational axioms" and
just plain definitions, which just amount to:
"when I speak of a "pandora sheaf", I'll just mean a sheaf with the following
[defining ] properties: (a) ... (b) .... ... (g) ... ".
In main-stream mathematics today, it seems to me that the most
comprehensive/"powerful" foundational axioms that will be
accepted as "unquestionable" by a typical mathematician are
the ones of ZFC (but some might prefer
von Neumann�Bernays�G�del set theory or NBG or other
set theories) ...
The whole list of the ZFC axioms and "axiom templates" a.k.a.
axiom schemas can be found here:
http://en.wikipedia.org/wiki/ZFC#The_axioms
I usually see the Axiom of Choice instead of
the Well-ordering Theorem in the axioms, but
it's simply a matter of tastes since
starting from ZF, the Axiom of Choice can be derived
from the Well-ordering Theorem and vice versa.
When using ZFC as the foundational theory
(sort of like the bedrock for developing maths),
the sets N and R (the real numbers), as well as
notions like functions, groups, topologies and so
on are all _defined_ words/properties in the same
sense as the "pandora sheaf" above could have
been defined by mathematicians who use sheaves.
The role of the foundational axioms, on the other
hand, is to permit deductions that almost all
mathematicians will agree on.
It's possible for "less ambitious" purposes to
use less abstract axiom systems such as
the axioms of Peano Arithmetic and develop
number theory from there. But the
going gets tough if one wanted to prove the
Prime Number Theorem (PNT) based on
Peano Arithmetic. This was done by Atle Selberg
and Erdos around 1950 but is reputedly very very
intricate. The PNT was first proved in the late 1800's
using complex analysis. Then there are
a number of known (finitistic combinatorial)
theorems that can be proven in ZFC, but that
provably are unprovable in plain Peano Arithmetic ...
Hope this helps,
David Bernier
Andrew Usher
> I made no claim that a larger model exists. Of course, if a larger
> model exists, then it would include objects which are not sets (in the
> same way that non-standard models of PA include objects which are not
> natural numbers).
I don't see these as analogous. We know already what a natural number
should be. A set, though, is an abstraction that includes the sort of
objects ZF deals with. So if ZFC includes any objects that are not
sets, it is inconsistent.
> You, on the other hand, are motivated by the view that there is a
> "real" set R and a "real" set N (and, one supposes, a real universe of
> sets containing R and N). Let us suppose so. I reckon that this
> universe is a model of ZFC -- do you disagree?
I don't know. I don't see any reason to believe one way or the other.
> >> Now, any function from N to R is a subset of N x R and hence an
> >> element of the universe of all sets. Thus, this same model has all
> >> possible functions N -> R.
>
> > But wouldn't this argument apply to all models, not only the standard
> > one. In that case, countable models could not exist (because N < R
> > would then be false in them), and you have acknowledged they do.
>
> No, because countable models do not include every function N -> R.
Why do countable models fail this test while the standard model
succeeds? Does every model with a 'really' uncountable R contain all
functions from its N to its R? Again, I don't know why I should
believe this.
Andrew Usher
Andrew Usher
> Andrew Usher <k_over_hb...@yahoo.com> writes:
> > Also, a theory that requires more assumptions is logically _weaker_.
>
> If you mean "axioms" where you write "assumptions", then, no, this is
> not how the terms "stronger" and "weaker" are used in logic.
OK. Your logic, anyway. I use it in the sense of Occam's razor.
> In the simplest case, where two theories T and T' share a common
> language, then T is stronger[1] than T' if, whenever T' |- P, then
> T |- P.
But ZFC can't share the same language with informal reasoning, since
the latter is at least second-order.
Andrew Usher
Jesse F. Hughes
> On Jan 15, 12:45 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>
>> I made no claim that a larger model exists. Of course, if a larger
>> model exists, then it would include objects which are not sets (in the
>> same way that non-standard models of PA include objects which are not
>> natural numbers).
>
> I don't see these as analogous. We know already what a natural number
> should be. A set, though, is an abstraction that includes the sort of
> objects ZF deals with. So if ZFC includes any objects that are not
> sets, it is inconsistent.
You're confused.
ZFC is a theory. It has various models. Those models may consist of
sets or of porkpie hats. ZFC is no different than N in this regard.
>> You, on the other hand, are motivated by the view that there is a
>> "real" set R and a "real" set N (and, one supposes, a real universe of
>> sets containing R and N). Let us suppose so. I reckon that this
>> universe is a model of ZFC -- do you disagree?
>
> I don't know. I don't see any reason to believe one way or the
> other.
You seem to believe that you have a notion of sets independent of
ZFC. Is this a reasonably well-developed notion? Can you not tell me
whether the universe of sets satisfies ZFC or not?
>> >> Now, any function from N to R is a subset of N x R and hence an
>> >> element of the universe of all sets. Thus, this same model has all
>> >> possible functions N -> R.
>>
>> > But wouldn't this argument apply to all models, not only the standard
>> > one. In that case, countable models could not exist (because N < R
>> > would then be false in them), and you have acknowledged they do.
>>
>> No, because countable models do not include every function N -> R.
>
> Why do countable models fail this test while the standard model
> succeeds? Does every model with a 'really' uncountable R contain all
> functions from its N to its R? Again, I don't know why I should
> believe this.
The standard model contains every set and hence every function
N -> R. A countable model does not contain every set and hence may
not (in fact, provably does not) contain every function N -> R.
I don't know why this is tricky to you, but I'm pretty tired of
repeating the same, dull observation.
Andrew Usher
> > I don't see these as analogous. We know already what a natural number
> > should be. A set, though, is an abstraction that includes the sort of
> > objects ZF deals with. So if ZFC includes any objects that are not
> > sets, it is inconsistent.
>
> You're confused.
>
> ZFC is a theory. It has various models. Those models may consist of
> sets or of porkpie hats. ZFC is no different than N in this regard.
- N is categorical and has only one model.
- What definition of 'set' are you using?
> >> You, on the other hand, are motivated by the view that there is a
> >> "real" set R and a "real" set N (and, one supposes, a real universe of
> >> sets containing R and N). Let us suppose so. I reckon that this
> >> universe is a model of ZFC -- do you disagree?
>
> > I don't know. I don't see any reason to believe one way or the
> > other.
>
> You seem to believe that you have a notion of sets independent of
> ZFC. Is this a reasonably well-developed notion? Can you not tell me
> whether the universe of sets satisfies ZFC or not?
Well, my definition is something like 'a definable, well-founded
collection of definable objects'. This is not a mathematical concept,
by intention.
> > Why do countable models fail this test while the standard model
> > succeeds? Does every model with a 'really' uncountable R contain all
> > functions from its N to its R? Again, I don't know why I should
> > believe this.
>
> The standard model contains every set and hence every function
> N -> R. A countable model does not contain every set and hence may
> not (in fact, provably does not) contain every function N -> R.
You are arguing in a circle. You are assuming that the standard model
is 'every set', and therefore your statement is a tautology. That is
my point.
Andrew Usher
Jesse F. Hughes
> On Jan 16, 5:47 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>
>> > I don't see these as analogous. We know already what a natural number
>> > should be. A set, though, is an abstraction that includes the sort of
>> > objects ZF deals with. So if ZFC includes any objects that are not
>> > sets, it is inconsistent.
>>
>> You're confused.
>>
>> ZFC is a theory. It has various models. Those models may consist of
>> sets or of porkpie hats. ZFC is no different than N in this regard.
>
> - N is categorical and has only one model.
The standard first order theory for N is not categorical, of course.
Were you speaking of some other theory?
In any case, categorical theories have only one model *up to
isomorphism*. They have infinitely many distinct models.
> - What definition of 'set' are you using?
The usual informal notion of set. What definition of natural number
are you using?
It's not really essential that I answer this question more clearly
than that, it seems to me. After all, I need not define "natural
number" in order to observe that non-standard models of PA include
elements which are not natural numbers.
>> >> You, on the other hand, are motivated by the view that there is a
>> >> "real" set R and a "real" set N (and, one supposes, a real universe of
>> >> sets containing R and N). Let us suppose so. I reckon that this
>> >> universe is a model of ZFC -- do you disagree?
>>
>> > I don't know. I don't see any reason to believe one way or the
>> > other.
>>
>> You seem to believe that you have a notion of sets independent of
>> ZFC. Is this a reasonably well-developed notion? Can you not tell me
>> whether the universe of sets satisfies ZFC or not?
>
> Well, my definition is something like 'a definable, well-founded
> collection of definable objects'. This is not a mathematical concept,
> by intention.
And you have no reason to believe that the universe of such
collections satisfies ZF or not? You don't know whether you can
construct unordered pairs of "definable, well-founded collections of
definable objects? You don't know if they are distinguished by their
members (i.e., extensionality holds)? And so on?
>> > Why do countable models fail this test while the standard model
>> > succeeds? Does every model with a 'really' uncountable R contain all
>> > functions from its N to its R? Again, I don't know why I should
>> > believe this.
>>
>> The standard model contains every set and hence every function
>> N -> R. A countable model does not contain every set and hence may
>> not (in fact, provably does not) contain every function N -> R.
>
> You are arguing in a circle. You are assuming that the standard
> model is 'every set', and therefore your statement is a
> tautology. That is my point.
The standard (i.e., intended) model *does* contain every set. What
could be more clear than that?
It seems clear to me that the universe of all sets is a model of ZF.
Is there any particular axiom of ZF that you doubt the universe of
sets satisfies?
--
Jesse F. Hughes
"Truth is common stuff, ready to your hand, but lies you have to make
yourself, and you can't be sure they are any good until you've
used them --- and then it's too late." John Steinbeck
Andrew Usher
> >> ZFC is a theory. It has various models. Those models may consist of
> >> sets or of porkpie hats. ZFC is no different than N in this regard.
>
> > - N is categorical and has only one model.
>
> The standard first order theory for N is not categorical, of course.
> Were you speaking of some other theory?
When we speak of N, we are referring to a particular set. As it turns
out, this set can be formalised in a second order theory first given
by Peano.
> > - What definition of 'set' are you using?
>
> The usual informal notion of set. What definition of natural number
> are you using?
So you agree that the informal notion of set has to precede any formal
set theory?
> > Well, my definition [of 'set'] is something like 'a definable, well-founded
> > collection of definable objects'. This is not a mathematical concept,
> > by intention.
>
> And you have no reason to believe that the universe of such
> collections satisfies ZF or not? You don't know whether you can
> construct unordered pairs of "definable, well-founded collections of
> definable objects? You don't know if they are distinguished by their
> members (i.e., extensionality holds)? And so on?
See below.
> > You are arguing in a circle. You are assuming that the standard
> > model is 'every set', and therefore your statement is a
> > tautology. That is my point.
>
> The standard (i.e., intended) model *does* contain every set. What
> could be more clear than that?
But (assuming there is such a model) how do we know which it is? You
just answered it: by appealing to our intuitive notions of what a set
should be in mathematics. Never mind the problems interpreting 'every
set' ...
> It seems clear to me that the universe of all sets is a model of ZF.
> Is there any particular axiom of ZF that you doubt the universe of
> sets satisfies?
Yes, the power set, at least. You might think that for every X, the
power set of X is definable, but that's not so: is the set of non-
trivial zeros of the zeta function included in the power set of {1/2 +
i*R} - That is equivalent to RH, and if RH is undecidable, so is that
power set. No doubt similar examples could be found for almost all
infinite power sets.
Besides, the set of all (other) sets is definable, yet we know that
taking its power set gives a contradiction.
Andrew Usher
David C. Ullrich
<k_over...@yahoo.com> wrote:
>On Jan 15, 7:28 am, David C. Ullrich <ullr...@math.okstate.edu> wrote:
>
>> >This is circular. Defining the 'real sets' in terms of ZFC makes no
>> >sense because, clearly, the 'real sets' do exist independently of any
>> >formal theory we may write.
>>
>> How in the world can you read the previous paragraph as a supposed
>> _definition_ of the "real sets"?
>
>It's the best we can do, given the inherent logical concept.
You seem to have trouble following things. Someone said something
which was not a definition of "the sets in ZFC". You objected that
his "definition of the sets in ZFC" was circular. When I point out
it was no such thing you reply that it's the best we can do?
Try to keep up.
MoeBlee
> On Jan 16, 5:47 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>
> > > I don't see these as analogous. We know already what a natural number
> > > should be. A set, though, is an abstraction that includes the sort of
> > > objects ZF deals with. So if ZFC includes any objects that are not
> > > sets, it is inconsistent.
>
> > You're confused.
>
> > ZFC is a theory. It has various models. Those models may consist of
> > sets or of porkpie hats. ZFC is no different than N in this regard.
>
> - N is categorical and has only one model.
I've not studied all the previous posts. That said, what do you mean
by 'N'? (For that matter, I don't know what Jesse means by 'N' in the
context above).
The notion of categoricity pertains to theories. So do you mean 'N' is
a certain theory? Do you mean (first order, from hereon unless
otherwise mentioned) PA? Or do you mean second order PA? Or do you
mean TH(N)? (which is the set of sentences true in the standard model
of PA)?
And no theory has just one model. Rather, some theories have just one
model TO WITHIN ISOMORPHISM (which is what 'categorcial' means). First
order PA is not categorical. Second order PA is categorical. TH(N) is
categorical. But both those categorical theories have more than one
model.
And a theory is not inconsistent merely for not being categorical.That
ZFC is not categorical doesn't in itself entail that ZFC is
inconsistent. Indeed, a theory with more than one model is perforce
consistent on account of having at least one model.
MoeBlee
Chris Menzel
said:
> ...
> And no theory has just one model. Rather, some theories have just one
> model TO WITHIN ISOMORPHISM (which is what 'categorcial' means). First
> order PA is not categorical. Second order PA is categorical. TH(N) is
> categorical.
First-order TH(N) isn't categorical, Moeb, in virtue of the upward L-S
theorem. (You knew that.)
Victor Porton
> What is so nice about the statement "ZFC (whatevuh) can prove
> everything provable." is:
There were much discussion on the question whether ZFC can prove
everything provable. But no one has given the answer.
Now I give the correct answer: ZFC cannot prove everything provable.
The question reformulated is: Can every math statement which we can be
sure proved in ZFC?
The answer is no, because we cannot prove "ZFC is consistent" in ZFC
but we are sure that ZFC is consistent.
Moreover we can construct a countable number of statements which we
may be sure are true but not provable in ZFC.
Let ZFC^1 = ZFC and ZFC^{n+1} is "ZFC^n and ZFC^n is consistent".
Oh, there are an intriguing question: Can we further extend the set of
mathematical statements which we can be sure in?
Aatu Koskensilta
> TH(N) is categorical.
Nope. As Chris notes we can see this by invoking the upward
L�wenheim-Skolem theorem. Perhaps more elementarily, the
non-categoricity of Th(N) follows also by compactness. (Like Chris, I
surmise you in fact know all this already...)
On an unrelated note, ZFC does not have a standard model the way PA
has. Rather, it has an intended interpretation: the quantifiers range
over sets, 'in' means membership. (ZFC does have many non-isomorphic
standard models on a technical definition of "standard model of set
theory".) When discussing the Skolem paradox we needn't concern
ourselves with such technical details: we may just as well consider
e.g. third-order arithmetic or Zermelo set theory, where the intended
interpretation can unproblematically (in ordinary mathematics) be given
by a model.
--
Aatu Koskensilta (aatu.kos...@uta.fi)
"Wovon man nicht sprechan kann, dar�ber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
Chris Menzel
<aatu.kos...@uta.fi> said:
> MoeBlee <jazz...@hotmail.com> writes:
>
>> TH(N) is categorical.
>
> Nope. As Chris notes we can see this by invoking the upward
> non-categoricity of Th(N) follows also by compactness.
Yes, that's an even better proof, as we can use compactness to show that
Th(N) isn't even aleph0-categorical.