On 7 Jan., 19:18, Zuhair <zaljo...@gmail.com> wrote:
> > In fact, it's consistent with ZF that there are sets x and y such that > > both |x| > |y| and |y| = |x|. > > True, but not in this extension of ZF! >
True, e.g., for indistinguishable reals. If you can't distinguish them, you cannot prove that they are more than the rationals. |R| = |Q|. But if you believe that they exists independently as different numbers (although nobody can prove it), then |R| > |Q|.