
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 >>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

