In an earlier [positing](https://mathoverflow.net/questions/413804/can-gch-fail-everywhere-every-way) to $\mathcal MO$, it appears that the answer to if the $\sf GCH$ can fail everywhere in *every way* is to the negative, this is the case in $\sf ZFC$, however it also appears that matters are more free in absence of $\sf AC$. So, here the quenstion is if the opposite direction in provable in $\sf ZF$.


The following question is about if it is true in $\sf ZF$ that for every stage $V\_\alpha$ of the cumulative hierarchy there is a cardinal $\kappa > |V\_\alpha|$ such that the cardinality of $V\_{\alpha+1} \neq \kappa$ in all models of $\sf ZF$?


Formally, this is:


$ \forall M: (M \models \mathsf {ZF}) \implies \\\forall \alpha \in M \, \exists \kappa \in M \big{(} (M \models |V\_\alpha| < \kappa) \land \forall N: N \approx M \to N \models|V\_{\alpha+1}| \neq \kappa \big{)}$


Where "||" refers to cardinality defined after Scott; $N \approx M$ means that $N$ is a model of $\sf ZF$ that shares the same ordinals and cardinals with $M$, i.e., $\operatorname {Ord}^M = \operatorname {Ord}^N$ and $\operatorname {Card}^M = \operatorname {Card}^N$


In the case of $\sf ZFC$, this seems to be the case, well at least for regular $\alpha$, and the singlular seems to be even more restricted. But, is this the case in $\sf ZF$?



Is existence of a cardinal that witness non-failure of GCH everywhere everyway, a theorem of ZF?
Here's a somewhat trivial answer.


Note that $V\_\alpha$, for an infinite $\alpha$, have a particularly nice set of properties which follow from the fact that $|V\_\alpha\times V\_\alpha|=|V\_\alpha|$.


Now, easily, if $\sf AC$ fails, we can find arbitrarily high such cardinals. Simply take $|V\_\alpha|+\aleph(V\_\alpha)$, where $\aleph(x)$ is the Hartogs number of $x$. By Tarski's lemma, this cardinal is not idemmultiple.


If $\sf AC$ holds, then refer to the previous question you've asked.


