Re: Skolem's Paradox and why is math the way it is?
From: J.E. (troubled6man_at_yahoo.com)
Date: 10/08/04
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Date: 8 Oct 2004 12:46:56 -0700
daryl@atc-nycorp.com (Daryl McCullough) wrote in message news:<ck624e02eb@drn.newsguy.com>...
> J.E. says...
>
> >What I am saying is that the first order ZF set theory axioms don't
> >prove the existence of uncountably many subsets of the naturals.
>
> Sure it does. In set theory, to say that a set S is uncountable is to
> say that there does not exist a bijection between that set and the
> naturals. ZF proves that there does not exist a bijection between P(N)
> and N.
>
> There is a distinction between
>
> A. There are uncountably many sets S such that ZF proves S exists.
> B. ZF proves that there are uncountably many sets S.
>
> There are only countably many sets that ZF can "name" (via an explicit
> definition). But ZF can prove that there are more sets than can be named.
What is the cardinality of the smallest sets that cannot be named? I
don't that there is REALLY an existance proof for sets that cannot be
proved to exist. I think you are using language in a sloppy manner.
Proving a lack of a bijection proves the class of bijections is small,
NOT that the class of subsets is large. Look at the countable model
to see this result. This REALLY isn't "falling off a log" kind of
obvious to you?
Your proof about a bijection just says that bijections can construct
elements that aren't in their image. But if the bijection does NOT
exist, then you have NOT proven that that element exists. You can't
ASSUME something incorrect and then get an intermediate result and
then get a contradiction and THEN pull out the intermediate result.
The PROOF of that additional element is FALLACIOUS. The proof is that
the bijection doesn't exist, not that the range of the non-existant
bijection is different than the power set. There is no proof of that
one way or another, it is independant of the ZF axioms.
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