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

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Jonathan Hoyle

Posts: 754
Registered: 12/10/04
Re: How many real numbers are there?
Posted: Jul 9, 2006 3:00 PM
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> 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.

That's true, but the converse need not be true. That is to say, there
exists sets which are not the union of some S_i.

> 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.

Incorrect. How did you get the "therefore"? Just because sets hold
for all the ZF axioms does *NOT* imply that they can be constructed
using your method. This is a major flaw in your thinking, and I am not
sure how you justify it even in your own mind.

> My set T has countably many elements.


> THEREFORE the cardinality of the reals is countable!!

Incorrect, for the reasons shown above.

> 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

This is where your step fails. Steps #4 & #5 are now moot.

> 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".

This is a very small issue. You are worrying about a minor crack here,
where a gaping hole exists at Step #3.

Jonathan Hoyle

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