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It is well known that working in the frame of $\text{ZF}$, the Generalized Continuum Hypothesis ($\text{GCH}$) implies the Axiom of Choice ($\text{AC}$), i.e. $\text{ZF}+\text{GCH}\vdash \text{AC}$.

But if we consider $\text{GCH}$ as a theory with ordinal many statements like $\text{GCH}=\{\text{CH}_{\alpha}~|~\alpha\in \text{Ord}\}$ such that $\text{CH}_{\alpha}$ is the statement $2^{\aleph_{\alpha}}=\aleph_{\alpha +1}$, then there is a natural question as follows:

Is assuming all of these strong statements really necessary to prove a weak proposition like Axiom of Choice?

Precisely:

Question (1): Is there a class $\text{C}\subsetneq \text{Ord}$ such that:

(a) The assumption $\{\text{CH}_{\alpha}~|~\alpha\in \text{C}\}$ is strictly weaker than the assumption $\{\text{CH}_{\alpha}~|~\alpha\in \text{Ord}\}$, i.e.

$\text{ZF}+\forall \alpha\in \text{C}~~~\text{CH}_{\alpha}\nvdash \forall \alpha\in \text{Ord}~~~\text{CH}_{\alpha}$

(b) The assumption $\{\text{CH}_{\alpha}~|~\alpha\in \text{C}\}$ is sufficient to prove $\text{AC}$, i.e.

$\text{ZF}+\forall \alpha\in \text{C}~~~\text{CH}_{\alpha}\vdash \text{AC}$

Question (2): If the answer of the question (1) is positive, can we choose $\text{C}$ to be a set not a proper class?

Question (3): What are the minimal classes (by inclusion order) like $\text{C}$ in the question (1)?

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  • $\begingroup$ What is $\{\mathrm{CH}_\alpha:\alpha\in\mathrm{Ord}\}$ actually? The language for set theory is countable, and so there cannot be ordinally many different sentences. Or are we assuming there is an intended (set) model of set theory at the back of our mind? $\endgroup$
    – Lawrence Wong
    Commented Nov 30, 2013 at 9:34
  • $\begingroup$ While the language is countable, we can talk about parameterizied formulas. Here $\alpha$ is a parameter. $\endgroup$
    – Asaf Karagila ♦
    Commented Nov 30, 2013 at 10:05
  • $\begingroup$ @Asaf How is this any different from specifying a single universally quantified formula? $\endgroup$
    – Adam Epstein
    Commented Nov 30, 2013 at 10:49
  • $\begingroup$ @Adam: It's not, really. But if you want to start talking about subclasses then it makes a difference. Moreover, you can say something like "There exists an ordinal such that ..." which specifies existence of ordinals satisfying the formula; but it doesn't explicitly points out which ordinals these are. $\endgroup$
    – Asaf Karagila ♦
    Commented Nov 30, 2013 at 11:18
  • $\begingroup$ @Asaf I think I am still confused. Could you give a syntactically explicit example illustrating this difference? $\endgroup$
    – Adam Epstein
    Commented Nov 30, 2013 at 11:52

2 Answers 2

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All you need for AC in the standard argument from GCH is that the GCH holds for an unbounded class of cardinals. The reason is that this is sufficient to conclude that any set of sets of ordinals is well-orderable, and this is sufficient to imply AC.

So the answer to question 1 is yes; any unbounded class $C$ suffices.

Meanwhile, merely knowing the GCH holds for cardinals below some cardinal is insufficient, since one can build the analogue of the symmetric models for $\neg\text{AC}$ above any cardinal, while preserving GCH below. Thus, there also can be no minimal $C$, since every sufficient $C$ is unbounded, and we may omit any proper initial segment of it and still have a sufficient $C$.

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    $\begingroup$ Does that "standard argument" for AC from GCH depend on the axiom of foundation? Is it true that, in the absence of the axiom of foundation, $2^{\aleph_{\alpha}}=\aleph_{\alpha+1}$ is not enough to prove AC, but you need the general GCH which says that there are no infinite cardinals $m$ and $n$ such that $m\lt n\lt2^m$? $\endgroup$
    – bof
    Commented Nov 30, 2013 at 4:52
  • $\begingroup$ I had in mind just the usual argument in ZF, which uses foundation. $\endgroup$
    – Joel David Hamkins
    Commented Nov 30, 2013 at 4:59
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    $\begingroup$ @bof It is indeed the case that the proof of AC from $(\forall\alpha)\,\text{CH}_\alpha$ requires the axiom of foundation. If you begin with a ZFC universe satisfying GCH and build a Fraenkel-Mostowski-Specker permutation model over it, then that model will still satisfy $(\forall\alpha)\,\text{CH}_\alpha$. (Specker is important here, to make the atoms violate foundation rather than extensionality.) $\endgroup$
    – Andreas Blass
    Commented Nov 30, 2013 at 5:37
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    $\begingroup$ @JoelDavidHamkins I always thought that, in choiceless set theory, the continuum hypothesis is not $2^{\aleph_0}=\aleph_1$ but the weaker assertion that there are no cardinals between $\aleph_0$ and $2^{\aleph_0}$; and the generalized continuum hypothesis is the same assertion for all infinite cardinals, not just alephs; and that "GCH implies AC" refers to an old Lindenbaum-Tarski-Sierpiński proof which does not require foundation. I wonder if I always had that wrong, or if I used to have it right and the fashions changed on me. (Oops, never mind--I just noticed Asaf Karagila's answer.) $\endgroup$
    – bof
    Commented Nov 30, 2013 at 11:25
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    $\begingroup$ @bof I'm not aware of any general convention as to what "GCH" should mean in the absence of AC. Of course, the issue comes up only in contexts (like the present one) where one can't take for granted that (just about any form of) GCH implies AC. So, in such situations, I would recommend not using "GCH" without some clarification of which version is intended. (The same recommendation applies to CH, even more strongly since CH doesn't imply AC.) $\endgroup$
    – Andreas Blass
    Commented Nov 30, 2013 at 14:38
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Let me add on Joel's answer and point out that in fact in $\sf ZF$ the following weakening of $\sf GCH$ holds:

For every $A$, if $A$ is well-orderable, then $\mathcal P(A)$ is well-orderable $\implies$ The Axiom of Choice.

From the above it is immediate that if there is a proper class of $\alpha$ such that $2^{\aleph_\alpha}=\aleph_{\alpha+1}$, then the axiom of choice holds. The proof is due to Herman Rubin.

One should note, however, that without the axiom of choice $\sf GCH$ can often be taken as "For every infinite set $A$, there is no $X$ such that $|A|<|X|<|\mathcal P(A)|$". This too implies the axiom of choice, and therefore the statement $\forall\alpha(2^{\aleph_\alpha}=\aleph_{\alpha+1})$.

The proofs that I know of the implications are different in nature. When assuming the weaker principle which is only for $\aleph$ numbers (or its weakening mentioned above), the proof usually goes by transfinite induction to show that every $V_\alpha$ is well-ordered and conclude the axiom of choice. When assuming that there is no intermediate cardinal between an infinite set and its power set, the proof usually goes to show that every set can be injected into its Hartogs number and therefore can be well-ordered.

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  • $\begingroup$ In your displayed implication, you want a quantifier over $A$. $\endgroup$
    – Joel David Hamkins
    Commented Nov 30, 2013 at 12:55
  • $\begingroup$ Dear Assaf, in "math.stackexchange.com/questions/200944/…" you have stated that Magidor has a list of open problems. Do you have them? $\endgroup$
    – Mohammad Golshani
    Commented Dec 1, 2013 at 9:28
  • $\begingroup$ @Mohammad: No, before he left to Fields last year I told him he should make such list, and he seemed positive about this idea, but he never really did it. Despite me reminding him several times since then. I'll see if I can find him on campus this week and ask him again. (Also, you keep writing Assaf, where my name is Asaf. :-)) $\endgroup$
    – Asaf Karagila ♦
    Commented Dec 1, 2013 at 9:34
  • $\begingroup$ Sorry for the mistake, $\endgroup$
    – Mohammad Golshani
    Commented Dec 1, 2013 at 9:38
  • $\begingroup$ @Mohammad: It's all fine. :-) $\endgroup$
    – Asaf Karagila ♦
    Commented Dec 1, 2013 at 9:42

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