```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
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