
Re: Inconsistency of the usual axioms of set theory
Posted:
Feb 18, 2009 10:42 AM


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.
Is it possible to express finiteness in ZF? Yes, because N (the set of natural numbers) is expressible in ZF. 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

