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:

>

>>(snip)

>>

>>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.

Stephen