About the relationship between the generalized continuum hypothesis and the axiom of choice

Question

I was trying to get a short, intuitive proof of Sierpinski’s theorem (gch implies axiom of choice) and I could but only by using the following gch2 for the generalized continuum hypothesis gch.

gch: Let S be an infinite set and S’ be the power set of S. If T is a subset of S’ such that |T| > |S|, i.e., T is not injectable into S, then |T| = |S’|, i.e., T and S’ are bijectable.

gch2($\alpha$): For ordinal $\alpha$ and V the cumulative hierarchy starting at the empty set, let S be a set where $\alpha$ is the least ordinal such that |S| $\leq$ |$V_\alpha$| (i.e., S is injectable into $V_\alpha$). Then |S| = |$V_\alpha$|.

gch2: gch2($\alpha$) is true if $\alpha$ is an infinite ordinal (of course it is false on all finite ordinals except 0, 1,and 2).

gch2 is easily proven to be equivalent to gch in ZF using Sierpinski’s theorem, but I couldn’t see how to get gch2($\lambda$) in ZF from gch without using Sierpinski’s theorem for the case where $\lambda$ is a limit ordinal.

gch2 asserts that every infinite ordinal stage of the cumulative hierarchy presents exactly one new cardinality that wasn’t presented at lower levels. The axiom of choice is sufficient to easily prove this for limit ordinal stages (usually expressed as the continuity of the beth function), but is it necessary? That is essentially my question presented below.

Two possibilities that I am interested in are: (1) proving in ZF that if gch2($\lambda$) holds for all limit ordinals $\lambda$, then the axiom of choice holds. Thus the axiom of choice would be equivalent to gch2($\lambda$) holding on all limit ordinals $\lambda$. And at the other extreme (2) proving in ZF that gch2($\lambda$) holds on all limit ordinals $\lambda$ (unconditionally).

My question is can either one of these possibilities be settled? If (1) can be done then the axiom of choice follows from gch2 immediately since all limit ordinals are infinite ordinals. It also suggests that Sierpinski has more to do with unions than power sets.

Without (1), gch2 easily implies trichotomy of cardinals, hence axiom of choice, in a manner not as short as (1) provides but pretty short and intuitive:

Let S and T be infinite sets. Then S and T are in the cumulative hierarchy by axiom of foundation. Let $\alpha$ and $\beta$ be the least ordinals such that |S| $\leq$ |$V_\alpha$| and |T| $\leq$ |$V_\beta$|, respectively. Since S and T are infinite sets, $\alpha$ and $\beta$ are infinite ordinals. By trichotomy of ordinals either $\alpha$ = $\beta$ or one of the two is greater than the other (say $\alpha$ > $\beta$). By gch2, which applies since $\alpha$ and $\beta$ are infinite ordinals, |S| = |$V_\alpha$| and |T| = |$V_\beta$|.

Now if $\alpha$ = $\beta$ then |S| = |T|. If $\alpha$ > $\beta$ then $V_\beta$ $\in$ $V_\alpha$, so that |$V_\beta$| < |$V_\alpha$| and thus |T| < |S|. Therefore trichotomy of cardinals is easily established in ZF from gch2.

Please note that the following two conditions are easily provably equivalent in ZF to gch2($\alpha$) without using Sierpinski, and are easily provably equivalent in ZF to gch if Sierpinski is used.

gch3($\alpha$): For $\alpha$ an ordinal, if S is a set such that |S| < |$V_\alpha$|, i.e., S is injectable into $V_\alpha$ but not bijectable, then |S| = |x| for some x $\in$ $V_\alpha$.

gch4($\alpha$): For $\alpha$ an ordinal, |$V_\alpha$| is the sup of the cardinalities |$V_\beta$| for $\beta$ < $\alpha$, in the sense that it is an upper bound with |$V_\alpha$| > |$V_\beta$|, for all $\beta$ < $\alpha$, and it is the least upper bound with |$V_\alpha$| $\leq$ any other such upper bound |S| of those cardinalities.

Similarly, note that the usual aleph formulations of gch, such as: for all ordinals $\alpha$, the power set of $\aleph_\alpha$ has cardinality $\aleph_{\alpha+1}$, are easily proved in ZF without Sierpinski to imply gch2, and therefore gch2 also provides short proofs of Sierpinski for these formulations of gch. These implications producing gch2 are true in these aleph formulations because the desired properties on both the limit ordinal and infinite successor ordinal stages of the cumulative hierarchy are easily derived from these formulations using the similar properties of the alephs.

So in summary, the generalized continuum hypothesis can be viewed as failing in ZF on all but the smallest finite ordinal stages of cumulative hierarchy, succeeding in ZFC on all limit ordinal stages, and is up for grabs in ZFC at the infinite successor ordinal stages. This is surprising to someone like me with just a pedestrian knowledge of set theory. We pedestrians know that the generalized continuum hypothesis fails at most finite stages but are unaware that it succeeds at so many infinite stages. Could this succeeding be used as an argument for accepting that hypothesis as a good condition for distinguishing finite and infinite sets, for those looking for such an argument? Also, possibility (1) implies that the generalized continuum hypothesis is what happens to the axiom of choice when the latter progresses (as the gch2 condition) to hold on all infinite ordinals instead of just the limit ordinals. Does that make accepting the generalized continuum hypothesis more palatable for those who already accept the axiom of choice?

Answer 1

Possibility (1) holds; i.e. ZF + "for all limit ordinals $\lambda$, GCH2($\lambda$) holds" implies choice. For it implies that for cofinally many ordinals $\lambda$, $V_\lambda$ can be wellordered. For fix an ordinal $\alpha$; then we can easily find a limit ordinal $\lambda>\alpha$ such that $\lambda$ does not inject into $V_\beta$ for any $\beta<\lambda$ (just define $\left<\lambda_n\right>_{n<\omega}$ by setting $\lambda_0=\alpha$, and given $\lambda_n$, $\lambda_{n+1}$ is the least $\lambda'>\lambda_n$ such that $\lambda'$ does not inject into $V_{\lambda_n}$; then set $\lambda=\sup_{n<\omega}\lambda_n$). But certainly $\lambda$ injects into $V_\lambda$. So we can apply GCH2($\lambda$) with $S=\lambda$ (since $\lambda$ is the least ordinal $\beta$ such that $S$ embeds into $V_\beta$). Therefore $|S|=|V_\lambda|$, i.e. $|\lambda|=|V_\lambda|$, i.e. there is a bijection between $\lambda$ and $V_\lambda$, so $V_\lambda$ is wellorderable, which suffices.