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