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I heard one can prove that GCH implies AC(the axiom of choice) in ZF. But I am confused with the meaning of GCH in ZF. Under AC, we can define cardinal exponentiation, so the cardinal 2 to the aleph is well defined. However, Without AC, How can we restate the GCH?

isab hud
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3 Answers3

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In ZF, the GCH can be stated as thus: for any ordinal $\alpha$, a bijection exists between $\omega_\alpha$ and $\beth_\alpha$. Here we can define$$\beth_0:=\omega,\,\beth_{\alpha+1}:=\mathcal{P}(\beth_\alpha),\,\beth_{\gamma}:=\bigcup_{\beta\in\gamma}\beth_\beta,$$where the last equation is for nonzero limit ordinals $\gamma$ only. All this can be done; it's certainly equivalent to the usual formulation. But the usual formulation uses alephs instead of ordinals.

J.G.
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  • Don't you mean between $\aleph_\alpha$ and $\beth_\alpha$? $\omega$ is usually an ordinal, at least in my book. Well, technically, we have $\omega_\alpha = \aleph_\alpha$, but it's about frame of mind. – Arthur Apr 27 '20 at 07:53
  • @Arthur I was deliberately using a notation where everyone knows which set it is, so we don't have to argue about what $\aleph_\alpha$ is. Even defining $\beth_\alpha$ in the way I have, rather than as "the cardinality of" something, had the same motivation. Since the GCH says certain bijections exist, we can state it without ever even discussing cardinalities. – J.G. Apr 27 '20 at 07:57
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    I should say that this is very unusual interpretation of GCH. While it is of course equivalent to the standard one, it is not usual to see it formulated like this (in the context of ZF). – Asaf Karagila Apr 27 '20 at 08:57
  • @AsafKaragila, the second definition of GCH in your answer implies this definition of GCH in the above answer? How do I prove... Help – isab hud Apr 27 '20 at 10:10
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    @isabhud: They are all equivalent, since they all imply the axiom of choice. – Asaf Karagila Apr 27 '20 at 10:13
  • @AsafKaragila To be fair, two inequivalent claims can imply AC, e.g. GCH & Zorn's lemma; in ZF, GCH is stronger than AC. But yes, I know what you mean. – J.G. Apr 27 '20 at 10:42
  • @J.G.: But these are not inequivalent... it's just that the proof of their equivalence is going through the fact that they both imply choice. This does not make the equivalence any less valid. – Asaf Karagila Apr 27 '20 at 11:39
  • @AsafKaragila Hence my last sentence. – J.G. Apr 27 '20 at 12:03
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You can formulate GCH in two ways:

  1. For every ordinal $\alpha$, $2^{\aleph_\alpha}=\aleph_{\alpha+1}$. Arguably, cardinals do not make a lot of sense outside of realm of ordinals (I am not advocate of this philosophy, though), and since this is the meaning of the statement in $\sf ZFC$, this should be the meaning in $\sf ZF$ as well.

  2. For every infinite set $x$, there is no set $y$ such that $|x|<|y|<|\mathcal P(x)|$. That is to say, there is no intermediate cardinalities between an infinite set and its power set.

Cantor's original formulation of CH was in line with (2), and only a decade later when the $\aleph$ notation came about that $2^{\aleph_0}=\aleph_1$ was stated.

It turns out that both of these are equivalent over $\sf ZF$, as they both imply the axiom of choice. So at the end we shouldn't worry about this too much. Nevertheless, from a local perspective, it is certainly possible that there are no intermediate cardinalities between $\omega$ and $\mathcal P(\omega)$, while at the same time $2^{\aleph_0}\neq\aleph_1$. This holds, for example, in Solovay's model.

Asaf Karagila
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In Sierpiński's book Cardinal and Ordinal Numbers, second edition revised, 1965, on p. 88 he defines it as follows:

The assumption that, no matter what an infinite set $A$ is like, there is no set which would be of greater power than $A$ and of less power than the set of all subsets of $A$ is called the generalized Continuum Hypothesis.

In short, $|A|\lt|X|\lt|\mathcal P(A)|$ never happens when $A$ is an infinite set. It is this form of the GCH that is meant in Sierpiński's theorem that the GCH implies the axiom of choice.

bof
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