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From: Jesse F. Hughes on 8 Jan 2010 09:40
scattered <still.scattered(a)gmail.com> writes:

> On Jan 5, 1:44 pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote:
>> Charlie-Boo <shymath...(a)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
From: David C. Ullrich on 8 Jan 2010 15:42
On Fri, 8 Jan 2010 05:51:02 -0800 (PST), Andrew Usher
<k_over_hbarc(a)yahoo.com> wrote:

>On Jan 8, 7:13�am, Charlie-Boo <shymath...(a)gmail.com> wrote:
>
>> > �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).

No, LS does not say that.

>Andrew Usher

From: Jesse F. Hughes on 8 Jan 2010 19:20
Andrew Usher <k_over_hbarc(a)yahoo.com> writes:

> On Jan 8, 8:31 am, "Jesse F. Hughes" <je...(a)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
"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.
From: David C. Ullrich on 9 Jan 2010 10:04
On Fri, 8 Jan 2010 14:46:42 -0800 (PST), Andrew Usher
<k_over_hbarc(a)yahoo.com> wrote:

>On Jan 8, 2:42 pm, David C. Ullrich <ullr...(a)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

From: Jesse F. Hughes on 11 Jan 2010 07:54
Andrew Usher <k_over_hbarc(a)yahoo.com> writes:

> On Jan 8, 6:20 pm, "Jesse F. Hughes" <je...(a)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
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