WM wrote:

> The power set axiom requires x⊆ ℕ <==> x ∈ P(ℕ). In words:
either a model of ZF contains all subsets of ℕ or it is not a model of ZF.

Then there wouldn't be a Skolem paradox, that ZF has countable models.
But one can more directly spot the fallacy in WMs argument. The fallacy
is based on that WM disregards tht a model also defines an universe

of discourse, and that in x⊆ ℕ <==> x ∈ P(ℕ) the variable x ranges over
the universe of discourse. So what "naively" are all subsets of N, by
appeal to intuition of some intended set universe,

is formally only a quantification of what the model offers, and we
cannot force a model to offer all subsets of N, by the axiom of
powerset alone. It creates the powerset only on the basis

what is available in the universe of discourse.

WM schrieb am Samstag, 4. November 2017 um 11:02:31 UTC+1:
> Hr e says about the axiom of power set: What 'goes wrong' is the rest of the axiom: y does not contain every subset of x; it only contains those subsets of x that are in M. So it is perfectly possible that this 'powerset' of x is countable. [Tim Chow: "Forcing for dummies", sci.math.research (10 Mar 2001)]
>
> Of course here he needs to address dummies. Everyone else would see through clearly: If a set x is in the model M, but a subset of x is missing in the model M, then at least one element of this x must be missing in the model M, then x is not in the model M
>
> In particular, If |N is in the model M, but a subset of |N is missing in the model M, then at least one element of |N must be missing in the model M, then |N is not in the model M, and M is not even a model of ZF for violating the axiom of infinity.
>
> Therefore the model M is uncountable from outside. A countable model of ZFC and a model of whole ZFC are mutually incompatible. Skolem's proof implies that ZFC cannot have any model.
>
> Regards, WM