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? > > Regards, WM
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. Of course, the universe or domain of discourse is generally assumed in naive set theories, and as well less naive set theories.
Feferman considers a lot whether the universe is the constructible universe. Levy I think of more in terms of acknowledgment of Skolem in Levy collapse, in a constructivist way.
