# Can GCH fail everywhere every way?

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: $$\forall\alpha>0:\beth_\alpha = \aleph_{f(\alpha)}$$

Formally:

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

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.

• does ZF prove the cofinality restriction without choice? [I think so, but too many years have passed ...] Jan 14, 2022 at 11:09
• 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. Jan 14, 2022 at 11:57
• Ah I see thanks. Jan 14, 2022 at 13:10
• 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. Jan 14, 2022 at 14:14
• @NameNo, thanks a lot! Jan 14, 2022 at 14:25

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 $$2^{\aleph_\omega} < \aleph_{\omega_4}$$.

• +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. 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.