
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 firstorder logic >> >> >> Theorem. Suppose ZermeloFraenkel set theory (ZF) be a firstorder >> >> >> 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 LowenheimSkolem theorem that it is >> >> >> not possible to characterize finiteness in firstorder logic [2], >> >> >> hence the part(ii) of the conditions above is not firstorder >> >> >> expressible. Consequently for every firstorder 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 socalled >> >> >"cranks," is attempting to use LowenheimSkolem to >> >> >prove ZF inconsistent. Those previous proof attempts >> >> >used _downward_ LS to construct a countable model >> >> >of ZF, yet R is uncountable, to claim contradiction. >> >> >> >But Hosseiny's proof attempts to use _upward_ LS to >> >> >obtain a nonstandard model of N, which he calls A, >> >> >so then if n is a nonstandard (infinite) natural: >> >> >> >n > n1 > n2 > n3 > ... >> >> >> >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 >> >> >LS 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 LowenheimSkolem theorem that it is >> >> not possible to characterize finiteness in firstorder logic [2], >> >> hence the part(ii) of the conditions above is not firstorder >> >> expressible." >> >> >> The problem is slightly subtle. It's true, in a sense, that >> >> >> (a) LS shows that it's impossible to characterize finiteness in >> >> firstorder logic. >> >> >> It's certainly true that ZF is a firstorder 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 postgrads." >> >> 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 LowenheimSkolem theorem that it is >> not possible to characterize finiteness in firstorder logic [2], >> hence the part(ii) of the conditions above is not firstorder >> 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 postgrads." >> 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 >firstorder >expressible. Hence, to establish ZF, you have to assume the existence >a set >which is not firstorder expressible. This contradicts the assumption >that ZF is >a firstorder theory. > >If you would like to have ZF as a firstorder 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 postgrads." in sci.logic.)

