From: kramsay@aol.commangled (Keith Ramsay) Subject: Re: How Many Transcendental Numbers? Date: 05 Mar 2001 02:13:07 GMT Newsgroups: sci.math Summary: Upper bounds on the order type of 2^{Aleph_0} In article , Steve Leibel writes: |Now the question is, which of the higher cardinalities describes the |real numbers? Is it the next one up from the countable ones, or is it a |higher cardinality than that? | |And the answer is -- nobody knows. And not only does nobody know, but |nobody knows in sort of a strange way. | |It turns out that you can just assume it's ANY ONE of the uncountable |cardinals (called the "alephs"). You can take as an axiom that the |number of reals is any aleph that you like, and you still get a |consistent theory and you can do conventional mathematics. This is not quite true. It's not possible, for example, for the continuum to be Aleph_omega, where omega is the smallest infinite ordinal. If I remember correctly, the continuum isn't the union of a countable collection of sets of smaller cardinality. A set of cardinality Aleph_omega is the union of subsets X1, X2, X3,... where Xn has cardinality Aleph_n. It's also not possible for the continuum to be Aleph_{c^+}, where c^+ is the smallest cardinality greater than c, or any other Aleph_x where x>c. The result uses some trick to convert a model M of set theory with an ordinal x which is not the limit of countably many smaller ordinals into a model M' in which Aleph_x is the continuum. Of course, if we do this trick to an ordinal x which is greater than the continuum in M, then in M' it's no longer true that x is greater than the continuum. Thus x may have lost its identity as c^+ or something else which might get moved by the trick. The result does imply, though, that for any integer n>0, the statement c=Aleph_n is consistent with ZFC, and so is c=Aleph_{omega+n}, and so on, which leaves the continuum problem pretty wide open. At different points in time, Goedel leaned toward c=Aleph_1 and c=Aleph_2 at least somewhat, although I don't think he ever was convinced of either one. Keith Ramsay