Models of ZF with more elements than ordinals Clearly a model of ZFC has as many ordinals as elements. Does this hold for models of ZF? In particular, if this holds for uncountable transitive models of ZF, then they are subject to Shoenfield absoluteness.
 A: To move this off the unanswered queue:
The $\mathsf{ZFC}$ situation is a bit subtle. For any $M\models\mathsf{ZFC}$ we have $\vert Ord^M\vert=\vert M\vert$. However, this may not be internal: there may be no $M$-definable class bijection $Ord^M\cong M$ (even using a parameter from $M$). Indeed, such a definable bijection exists iff we have $M$ satisfies global choice or equivalently $M\models \exists x(V=HOD(x))$.
For $\mathsf{ZF}$ even the external equipollence fails. In fact we have the following extraordinary situation:

(Friedman): Suppose $M$ is a countable transitive model of $\mathsf{ZF}$. Then there is a transitive $N\models\mathsf{ZF}$ with $Ord^M=Ord^N$ and $M\subseteq N$ such that $\vert N\vert=$ $\beth_{Ord^M}$.

This leaves open the question of how extremely we can have $\vert M\vert>\vert Ord^M\vert$ for uncountable-height transitive (or even nontransitive) $M\models\mathsf{ZF}$. I have to assume the situation isn't any better, but I don't have an immediate proof.

That said, although usually stated for $\mathsf{ZF+DC}$ Shoenfield absoluteness in fact holds for $\mathsf{ZF}$ models; see this answer of Dorais. The point is that we don't need full Mostowski absoluteness in the proof of Shoenfield.
