I'm looking through Set theory and the continuum hypothesis by Smullyan and Fitting. In chapter 9 they give a nice crash course in cardinality. They define the cardinals as the ordinals that can't be put into one-to-one correspondence with any lesser ordinal and they define the cardinality of a well-orderable set as the unique cardinal it can be put into one-to-one correspondence with. The prospect of assigning cardinality to non-well-orderable sets is not discussed (I'm vaguely aware of Scott's trick, but that's not what my question is about.)
Toward the end of that same chapter, at the beginning of section 9, they state the GCH as "if $x\preceq y \preceq \mathcal{P}(x)$ then either $x \cong y$ or $y\cong \mathcal{P}(x).$" There's no assumption that the sets in question need to be well-orderable. They prove that this statement of the GCH => axiom of choice.
However, at the beginning of chapter 15, where they are going to prove that the GCH holds in the constructible universe $L,$ they state the GCH as "for any infinite cardinal $c$ the cardinality of $\mathcal{P}(c)$ is the smallest cardinal greater than $c$".
It's obvious that these two statements of GCH are equivalent under the axiom of choice, but not as much without it, since the second statement is superficially weaker doesn't have anything to say about non-well-orderable sets (and it seems to take it for granted that the power set of a cardinal is well-orderable).
This answer to a similar question suggests that the difference is substantial (although it addresses the regular CH and it's conceivable that since the GCH is stronger the weaker version actually proves the stronger one).
Once they've proved that the second version of GCH is consistent w/ ZF they conclude by saying that the axiom of choice is consistent with ZF because GCH => AC is a theorem of ZF. Of course, they spare us the gory details of showing that the proof can be formalized, but I'm left wondering if we even have an informal proof of this fact, since the proof they gave that GCH => AC relied on their first (ostensibly) stronger version of GCH that applies to non-well-orderable sets.
This doesn't seem to be a major issue since they indicate shortly afterward that since $L$ is well-ordered, the consistency of AC is much easier to to prove on its own than as a consequence of GCH. So you could presumably just prove AC over $L$, and then prove that AC implies the weak and strong GCH are equivalent, and then you can prove the weak GCH and have everything you want. Still I'm curious whether they've made an oversight here or if I'm thinking about it wrong.
So with that background, I'll condense it to three questions:
- Is their strong version of GCH equivalent to their weak one? (W/O AC, of course)
- If not, does the weaker one still imply the axiom of choice?
- If not, are Smullyan and Fitting making a minor error when they say proving (the weaker version of) GCH consistent w/ ZF and then proving GCH => AC over ZF is a viable route to proving AC is consistent with ZF? Or have I thought this through wrong?
(While I have my logician hat on, I should say that it's possible that it is a viable route, but they've nonetheless made an error; whether I've thought it through wrong in independent of those two things.)
