Date: Jul 7, 2006 11:43 PM
Author: Stephen Montgomery-Smith
Subject: Re: How many real numbers are there?

Mike Deeth wrote:
> Eric Schmidt wrote:

>>You have not proved that every set (or even every real number) is in
>>some S_i. In fact, you yourself construct a set not in any S_i, namely
>>T. Therefore you have not found any flaw in set theory.

> Of course my T does not contain every set, that would be really silly.
> However, the elements of my set T satisfy the ZF axioms. They do this
> by very construction, and you cannot argue about that! For example, if
> x and y are in my set T, then so is x union y, {x,y}, etc.
> The construction of the real numbers does not depend on the specific
> details of the set theory. *** All it requires is that ZF axioms hold,
> and they DO hold for elements of my set T *** Therefore, we can
> construct the "reals" (ie, a complete ordered field) using nothing but
> the elements of my set T. My set T has countably many elements.
> THEREFORE the cardinality of the reals is countable!!
> If you want to dispute this, please consider the following string of
> statements and tell me which one is false. Just one will suffice.
> 1. The construction of the reals (ie, a complete ordered field) does
> not depend on particular details of one's set theory, only on the fact
> that the ZF axioms hold
> 2. ZF axioms hold for elements of my set T
> 3. We can construct, in the usual way, the "reals" (ie, a complete
> ordered field), in such a way that each "real" is an element of my set
> T
> 4. My set T has countable cardinality
> 5. THEREFORE, the "reals" (ie, one particular complete ordered field,
> namely the one in statement 3) have countable cardinality.
> Note that in making the deduction in step 5 we use the lemma "A subset
> of a countable set is countable". As some point out, this might take a
> little more power than ZF. But this is fine! My set T was constructed
> in ZFC! The *elements* of my set T may or may not satisfy a choice
> axiom. The complete ordered field constructed in statement 3 consists
> of elements of T, and each element of T is a set in ZFC, and therefore
> in the original ZFC I started out with, the complete ordered field of
> statement 3 is a subset of the countable set T, and thus is countable.
> The *very most* the evil Cantorians can salvage at this point would be
> a statement like, "the assertion that the reals are uncountable, like
> the axiom of choice or continuum hypothesis, is an independent
> assertion and can't be proved or disproved".
> I hope you're having a wonderful day,
> Nathan the Great
> Age 11 :-)

I think your construction can also be used to show that there are only
countably many sets - you don't have to limit yourself to real numbers
here. I think that you are trying to get to a proof that ZF or ZFC has
a countable model.

Or to put it another way, I think your argument fails at point (3),
because you can only show that T is complete in the sense that every
constructable subset of T has a least upper bound in T - you cannot show
this for every subset of T.