TITLE: Can GCH fail everywhere every way?
QUESTION [10 upvotes]: 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.

REPLY [11 votes]: 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.