|
|
Re: Inconsistency of the usual axioms of set theory
Posted:
Feb 19, 2009 7:47 AM
|
|
On Wed, 18 Feb 2009 07:42:58 -0800 (PST), Student of Math
<omar.hosseiny@gmail.com> wrote:
>On Feb 17, 2:29 pm, David C. Ullrich <dullr...@sprynet.com> wrote:
>> On Mon, 16 Feb 2009 02:58:04 -0800 (PST), Student of Math
>>
>>
>>
>>
>>
>> <omar.hosse...@gmail.com> wrote:
>> >On Feb 13, 1:27 pm, David C. Ullrich <dullr...@sprynet.com> wrote:
>> >> On Thu, 12 Feb 2009 19:48:04 -0800 (PST), lwal...@lausd.net wrote:
>> >> >On Feb 12, 11:27 am, Student of Math <omar.hosse...@gmail.com> wrote:
>> >> >> I was trying to find the answer for some problems like Goldbach
>> >> >> Conjecture and continuum problem, but I found that the axiomatic
>> >> >> system for which these problems are discussed in it (ZF), is
>> >> >> inconsistent.
>>
>> >> >Let me warn you that those who attempt to prove ZF
>> >> >inconsistent are usually labeled "cranks," "trolls,"
>> >> >and much worse names.
>>
>> >> There's an important distinction that you _insist_ on
>> >> missing, no matter how many times it's pointed out
>> >> to you. The following are very different:
>>
>> >> (i) Attempting to prove ZF inconsistent.
>>
>> >> (ii) Stating that one _has_ proved ZF inconsistent,
>> >> and then ignoring or misunderstanding simple
>> >> explanations of why the proof is simply wrong.
>>
>> >> >Also, threads discussing the
>> >> >consistency of ZF end up being be hundreds of posts
>> >> >in length. Indeed, I'm surprised that this thread
>> >> >doesn't already 100 posts...
>>
>> >> >> ----------------------------------------------
>> >> >> Inconsistency of ZF in first-order logic
>> >> >> Theorem. Suppose Zermelo-Fraenkel set theory (ZF) be a first-order
>> >> >> theory, then it is inconsistent.
>> >> >> Proof.
>> >> >> Whereas the axiom of foundation eliminates sets with an infinite
>> >> >> descending chain each a member of the next [1], we show that the
>> >> >> existence of those sets is postulated by axiom of infinity.
>> >> >> According to the axiom of infinity there exists a set $A$ such
>> >> >> that $\emptyset\in A$ and for all $x$, if $x\in A$, then
>> >> >> $x\cup\{x\}=S(x)\in A$ [1]. Note that $x\in S(x)$, let $B\subset
>> >> >> A$ such that $x\in B$ if and only if there is no $y\in A$ with
>> >> >> $S(y)=x$. Assuming that for all given sets the descending chain
>> >> >> each a member of the next is finite, we may arrange our ideas as
>> >> >> follows
>> >> >> $A$ is a set such that $B\subset A$ and for all $x$, if $x\in A$,
>> >> >> then
>> >> >> (i) $S(x)\in A$, and
>> >> >> (ii)$x$ is obtained from an element of $B$ by a finite
>> >> >> step(s).
>> >> >> Where a "step" is to obtain $S(x)$ from $x$. This may be
>> >> >> written more precisely
>> >> >> \begin{equation}A=\bigcup_{z\in B}{\Omega_{z}}\end{equation}
>> >> >> where $\Omega_{z}=\{z,S(z),S(S(z)),\ldots \}$. One may define
>> >> >> $\Omega_{z}$ as a set which satisfies both (i) and (ii), with
>> >> >> $B=\{z\}$.
>> >> >> It follows from the upward Lowenheim-Skolem theorem that it is
>> >> >> not possible to characterize finiteness in first-order logic [2],
>> >> >> hence the part(ii) of the conditions above is not first-order
>> >> >> expressible. Consequently for every first-order expressible set
>> >> >> which its existence is asserted by axiom of infinity (condition (i)),
>> >> >> the condition (ii) does not holds. This implies the existence a set
>> >> >> with an infinite descending chain each a member of the next, a
>> >> >> contradiction.
>>
>> >> >We observe that Hosseiny, like a few other so-called
>> >> >"cranks," is attempting to use Lowenheim-Skolem to
>> >> >prove ZF inconsistent. Those previous proof attempts
>> >> >used _downward_ L-S to construct a countable model
>> >> >of ZF, yet R is uncountable, to claim contradiction.
>>
>> >> >But Hosseiny's proof attempts to use _upward_ L-S to
>> >> >obtain a nonstandard model of N, which he calls A,
>> >> >so then if n is a nonstandard (infinite) natural:
>>
>> >> >n > n-1 > n-2 > n-3 > ...
>>
>> >> >is a infinitely descending chain, once again to
>> >> >claim contradiction and hence ZF inconsistent.
>>
>> >> >I know that the "standard" mathematicians (i.e., the
>> >> >ones who support ZF(C)) have attacked the downward
>> >> >L-S proof, and they'll likely invalidate Hosseiny's
>> >> >proof as well.
>>
>> >> >In the downward proof, it was pointed out that the
>> >> >model may be countable, but from the perspective of
>> >> >that model, R is still uncountable. Similarly, I
>> >> >suspect that the descending chain is infinite, but
>> >> >it is finite from the perspective of this model.
>>
>> >> >The defenders of ZF(C) will probably explain it
>> >> >more eloquently than I can. Soon, there'll be
>> >> >hundreds of posts all showing Hosseiny's why his
>> >> >proof won't be accepted as a proof of ~Con(ZF).
>>
>> >> Erm, the reason it won't be accepted is that it's
>> >> simply _wrong_. Is there some reason incorrect
>> >> proofs _should_ be accepted?
>>
>> >> There's at least one major misunderstanding
>> >> that jumps out:
>>
>> >> "It follows from the upward Lowenheim-Skolem theorem that it is
>> >> not possible to characterize finiteness in first-order logic [2],
>> >> hence the part(ii) of the conditions above is not first-order
>> >> expressible."
>>
>> >> The problem is slightly subtle. It's true, in a sense, that
>>
>> >> (a) LS shows that it's impossible to characterize finiteness in
>> >> first-order logic.
>>
>> >> It's certainly true that ZF is a first-order theory. Nonetheless
>> >> it's also true that
>>
>> >> (b) "S is finite" is expressible in the language of ZF.
>>
>> >> The reason that (a) and (b) do not contradict each other
>> >> has to do with the disctinction between "finite" and
>> >> "finite according to this model". Much like the explanation
>> >> for the traditional LS "paradox", that ZF, if consistent,
>> >> must have a countable model, although ZF proves
>> >> that there are more than countably many sets.
>>
>> >> Hmm. Come to think of it, looking at what you wrote
>> >> you seem to have realized all by yourself that this is
>> >> a problem. So what's your point?
>>
>> >> 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.)- Hide quoted text -
>>
>> >> - Show quoted text -
>>
>> >It does not concern my argument that "finiteness is expressible in
>> >ZF",
>>
>> Huh? It certainly does matter. You said
>>
>> (ii)$x$ is obtained from an element of $B$ by a finite
>> step(s).
>>
>> and then in explaining why this supposedly led to a
>> contradiction you said
>>
>> "It follows from the upward Lowenheim-Skolem theorem that it is
>> not possible to characterize finiteness in first-order logic [2],
>> hence the part(ii) of the conditions above is not first-order
>> expressible."
>>
>> Condition (ii) _is_ expressible in ZFC, hence your argument
>> is simply wrong.
>>
>> >as if ZF may be inconsistent then it is possible to express
>> >every thing in it.
>>
>> That's simply not so.
>>
>> >If you have proven that ZF is consistent.
>>
>> That's not a sentence, so I'm not sure what the appropriate
>> reply is. But I certainly have not proven that ZFC is
>> consistent - I never claimed to have done so, and
>> nothing I've said requires that I have done so.
>>
>> >As I said before, you have assumed a sentence S be true and the you
>> >discuss if it may be true or false.
>>
>> Yes, you've said this before. That doesn't make it true.
>>
>> 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.)- Hide quoted text -
>>
>> - Show quoted text -
>Hello,
>I can explain my ideas as more as you want.
You'd be better off trying to understand the explanations
of why your proof is wrong.
>Is it possible to express finiteness in ZF?
Yes.
>Yes, because N (the set of natural numbers) is expressible in ZF.
No, we could express finiteness even without N.
>Why the set N is expressible in ZF?
>Because of the axioms of ZF.
>
>the axiom of infinity states
>
>There exists a set A such that 0 is in A and for all x, if x is in A,
>then
>xU{x}=S(x) is in A.
>
>I showed that the set A which its exsitence is asserted by axiom of
>infinity, and
>it has no elements with an infinite descending chain each a member of
>the next,
>is of the form ($$A=\bigcup_{z\in B}{\Omega_{z}}$$), which is not
>first-order
>expressible. Hence, to establish ZF, you have to assume the existence
>a set
>which is not first-order expressible. This contradicts the assumption
>that ZF is
>a first-order theory.
>
>If you would like to have ZF as a first-order theory, then the set A
>which its
>existence is asserted by axiom of infinity has an infinite descending
>chain each
>a member of the next, a contradiction.
>
>Regards,
>Omar hosseiny
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.)
|
|
