On 2 Jul., 09:06, "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote: > On Jul 2, 12:01 am, WM <mueck...@rz.fh-augsburg.de> wrote: > > > On 2 Jul., 04:03, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: > > > > My answer to your question above is similar. The question whether a countable > > > union of countably many sets is countable is implied by the axiom of countable > > > choice, which is not provable from ZF. So negating the axiom of countable > > > choice (which implies negating the axiom of choice) we can have situations > > > where the conclusion is false. > > > Feferman and Levy showed that the statement that the set of all real > > numbers is the union of a denumerable set of denumerable sets cannot > > be refuted. > > > So does it depend on the chosen axioms, whether R is countable?
> > No, countable choice is a theorem of ZF. It is attributed to > Hausdorff "countable unions of countable sets are not necessarily > dissimilar from uncountable sets". However, to him is attributed a > more direct statement. > > R can always be considered countable in for example some constructible > universe containing the real numbers via Skolemization. That's where > it's a countable model, even if R is still uncountable. The model is > supposed to be some big ordinal, bigger than any it models, but there > is no such thing in regular set theories.
There is no model of ZFC. My question is: Is the set of real numbers uncountable in ZF, is it uncountale in ZFC. Is a countable union of contable sets uncountable in ZF, is it uncountable in ZFC?
That boils down to the question: Does Cantor's proof show the same as Feferman and Levy showed? Or does the result depend on the chosen axioms?