On Jul 2, 9:28 am, MoeBlee <jazzm...@hotmail.com> wrote: > On Jul 2, 12:06 am, "Ross A. Finlayson" <ross.finlay...@gmail.com> > wrote: > > > countable choice is a theorem of ZF. > > No, it's not. > > MoeBlee
The definition of countable is there existing a bijective function between some set and the natural integers or here their ordinals. A choice function is representable by any such function bijecting an ordinal as a set to a set. Then, where countable choice means there are choice functions for countable sets, otherwise there are "countable" sets with no provable bijection to the naturals, where if that's undecideable then so is whether they're countable, not disproving the definition.