# Can we prove AC from the statement “There is no $\aleph$ cardinal strictly between $\operatorname{CARD}(X)$ and $\operatorname{CARD}(2^X)$”?

If $X$ is a set, let $\operatorname{CARD}(X)$ denote the Cardinal number of $X$. Let GCH(1) be the statement "If $K$ is an infinite initial ordinal number, then there exists no initial ordinal number $L$ such that $\operatorname{CARD}(X)<\operatorname{CARD}(L)<\operatorname{CARD}(2^X)$". Let GCH(2) be the stronger statement "If $X$ is an infinite set, then there exists no set $Y$ such that $\operatorname{CARD}(X)<\operatorname{CARD}(Y)<\operatorname{CARD}(2^X)$". Most discussions of GCH seem to be about GCH(1). Assume that we are working in ZF-without the Axiom of Choice. Sierpinski proved that GCH(2) implies the Axiom of Choice. Does GCH(1) also imply the Axiom of Choice?

• Most of my question somehow got deleted. Can this be fixed? – Garabed Gulbenkian May 7 '15 at 18:34
• There's an edit link just below the tag, click on it and rewrite what was lost. – Henrik supports the community May 7 '15 at 18:46
• So basically,GCH(2) says that no cardinal number lives strictly between $|X|$ and $2^{|X|}$, while GCH(1) says that no aleph number lives there. Is that right? – goblin May 7 '15 at 19:05
• Yes. The sets which occur in GCH(1) are all Alephs – Garabed Gulbenkian May 7 '15 at 19:27
• Related: mathoverflow.net/q/206090/7206 – Asaf Karagila May 8 '15 at 19:30

As it turns out, the answer is negative. But the solution, as far as I understand requires [very] large cardinals.

To see this, first note that the first statement necessarily means that $X$ can be well-ordered (since it can be mapped into a well-ordered set injectively). So it suffices to show that for well-ordered sets this happens, while the axiom of choice fails. In fact it suffices to show that for every ordinal $\alpha$, $\aleph_{\alpha+1}\nleq2^{\aleph_\alpha}$.

Apter showed that it is consistent from a $3$-huge cardinal that there is a model of $\sf ZF$ in which every successor cardinal is Ramsey. (The paper mentions at the end that the consistency strength can be lowered to an almost huge cardinal.)

The proof that Ramsey cardinals are strongly inaccessible can be followed in $\sf ZF$ to obtain that if $\kappa\leq2^\lambda$ for some $\lambda<\kappa$, then $\kappa$ is not weakly compact, and in particular not Ramsey. The idea lies in the fact that in the lexicographic ordering of $2^\lambda$ there is no sequence which is strictly increasing, or strictly decreasing, of length $\lambda^+$.

While the failure of $\sf GCH(1)\rightarrow AC$ is probably not equiconsistent with a $3$-huge, or even an almost huge cardinal, it does imply the consistency of some large cardinals, to say the least.

1. If $\kappa^+\nleq2^\kappa$, then given any inner model $M$ which satisfy $\sf AC$, it is necessarily the case that $(\kappa^+)^M<(\kappa^+)^V$. Otherwise the two models agree on $\kappa^+$, and in $V$ the set $\mathcal P(\kappa)$ can only be larger.

This means that in any inner model satisfying choice, $\kappa^+$ is a limit cardinal. If $\kappa^+$ is regular there, it is inaccessible; if $\kappa$ happened to be singular, then we move to the next point.

2. Suppose that $\kappa$ is singular in $V$, either it is regular in some inner model, in which case $0^\#$ exists and we jumped quite high in the consistency strength; or $\kappa^+$ is a limit cardinal in $L$ (by the argument above) and therefore $(\kappa^+)^L\neq(\kappa^+)^V$ and we get back to $0^\#$.

My guess is that you can obtain this for successor ordinals without much more than a proper class of inaccessible cardinals; but once you need to change the successors of singular cardinals, you are in for much larger cardinals than just that.

• Would the answer still be negative if the large cardinal axiom were inconsistent with ZF? – Garabed Gulbenkian May 8 '15 at 18:02
• I don't know. I believe that you can probably show that at least some large cardinals are needed. And my guess is that probably "some" is going to be relatively large cardinals at least (although my current estimate is much weaker than the axioms used by Apter. I don't know enough to fully back my claims, though, they are just educated guesses). – Asaf Karagila May 8 '15 at 18:18
• Thanks for all this information involving areas of set theory that I know almost nothing about. I guess I really opened a can of worms with this question. – Garabed Gulbenkian May 9 '15 at 15:16
• Well, it's at least interesting to me. :-) – Asaf Karagila May 9 '15 at 15:48