The following question is about if it is compatible to add to $\sf ZF$ an axiom asserting the existence of a countable transitive model of $\sf ZF$ such that for every strictly increasing function $f$ on the ordinals, we have a transitive countable model of $\sf ZF $ satisfying: $$\forall\alpha>0:\beth_\alpha = \aleph_{f(\alpha)}$$


$ \exists M: M \equiv \operatorname {CTM}(\mathsf {ZF}) \land \forall f \subseteq M \ \big{(}\\f: \operatorname {Ord}^M \to \operatorname {Ord}^M \land \forall \alpha \forall \beta \, ( \beta > \alpha \to f(\beta) > f(\alpha) ) \\ \implies \\ \exists N : N \equiv \operatorname {CTM}(\mathsf {ZF}) \land \operatorname {Ord}^N =\operatorname {Ord}^M \land (N \models \forall \alpha > 0 : \beth_\alpha=\aleph_{f(\alpha)})\big{)} $

Where "$\equiv\operatorname {CTM}(\mathsf {ZF})$" means "is a countable transitive model of ZF"

So this is to say that the generalized continuum hypothesis can fail everywhere and in everyway.

  • 1
    $\begingroup$ does ZF prove the cofinality restriction without choice? [I think so, but too many years have passed ...] $\endgroup$
    – NameNo
    Jan 14, 2022 at 11:09
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    $\begingroup$ The fact that for each ordinal $\alpha$, $\beth_\alpha$ is well defined as a well-ordered cardinal number, is by itself equivalent to AC. Thus, $N$ is really a model of ZFC. $\endgroup$ Jan 14, 2022 at 11:57
  • $\begingroup$ Ah I see thanks. $\endgroup$ Jan 14, 2022 at 13:10
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    $\begingroup$ My point (see mathoverflow.net/questions/78627/… ) is that also any other formalizaition of "fail everywhere in every way" is killed since restrictions do exist. $\endgroup$
    – NameNo
    Jan 14, 2022 at 14:14
  • $\begingroup$ @NameNo, thanks a lot! $\endgroup$ Jan 14, 2022 at 14:25

2 Answers 2


No. An early nontrivial constraint on the $\beth$ function comes from Kőnig's Theorem, that for all infinite $\kappa$, $\mathrm{cf}(2^\kappa)>\kappa$. This implies that we cannot have $\beth_\alpha = \aleph_{f(\alpha)}$ for all $\alpha$, when $f(1) = \omega$, nor when $f(\omega+1)$ is a cardinal below $\aleph_\omega$.

Another constraint is Silver's Theorem, that if GCH holds below a singular of uncountable cofinality, then it holds at that singular as well.

Other constraints come from Shelah's PCF theory. Shelah showed that if $\aleph_\omega$ is a strong limit, then $2^{\aleph_\omega} < \aleph_{\omega_4}$.

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    $\begingroup$ +1. To the OP, it's worth explicitly saying that, per Easton's theorem, Konig's theorem is the only real restriction on the behavior of the continuum function ... on regular cardinals; as Silver's theorem and PCF theory show, the situation on singular cardinals is much more intricate. $\endgroup$ Jan 14, 2022 at 17:05

When working in ZF, one can have more freedom. See An Easton-like Theorem for Zermelo-Fraenkel Set Theory with the Axiom of Dependent Choice and An Easton-like theorem for Zermelo-Fraenkel Set Theory without Choice.


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