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Topic: How many real numbers are there?
Replies: 18   Last Post: Jul 9, 2006 3:00 PM

 Messages: [ Previous | Next ]
 Stephen Montgomery-Smith Posts: 2,351 Registered: 12/6/04
Re: How many real numbers are there?
Posted: Jul 7, 2006 11:43 PM

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
>>

>
>
> 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

Date Subject Author
7/7/06 Mike Deeth
7/7/06 Nathan
7/7/06 Toni Lassila
7/7/06 Nathan
7/8/06 Denis Feldmann
7/8/06 Aatu Koskensilta
7/7/06 Stephen Montgomery-Smith
7/7/06 Anonym1723
7/7/06 Jonathan Hoyle
7/7/06 Mike Deeth
7/7/06 Eric Schmidt
7/7/06 Mike Deeth
7/7/06 Stephen Montgomery-Smith
7/9/06 Jonathan Hoyle
7/8/06 Jonathan Hoyle
7/9/06 Anonym1723
7/8/06 Virgil
7/7/06 mensanator
7/8/06 donstockbauer@hotmail.com