> Am Donnerstag, 9. November 2017 10:25:08 UTC+1 schrieb Alan Smaill: > >> >> If you want to say that is equivalent, *you* need to >> >> justify that equivalence. >> > >> > On the contrary. Everybody who denies that the power set is >> > uncountable has to show at least one subset of Z that is not a set. >> >> More diversionary tactics. >> >> Try and justify your own claim: the burden of proof is with you. > > Done already. Did you miss it? Here it is again: Assume there was a > subset of Z that is not an element of the power set of Z (in the > model) but is an element of the power set of Z (in our "universe"). > > Contradiction,
> because the axioms of ZF do not refer to models. Either > a model satisfies all of them (and here all means all) or it is not a > model.
Repeating the bleeding obvious that no-one has ever disputed.
> In particular there is no difference with respect to this point > between "our universe" and any other model.
Except, of course, that there are many different models!
To remind you, you claim that *all possible combinations* of sets appearing in |N must appear in the power set.
The axioms do not say this, you carefully avoid this question each time you are asked to give a justification.