I wish to know which $V_\alpha$ contain cardinalities of all its elements (for an ordinal $\alpha$).
I've conjectured that all of them, and tried to prove this by the transfinite induction on $\alpha$.
$V_0$ contains no elements, so that's true by default.
Assume that $\forall \beta < \alpha \ \forall A \in V_\beta, |A| \in V_\beta$ for some limit ordinal $\alpha$. Then, if $A \in V_\alpha = \bigcup_{\beta < \alpha} V_\beta$, then there is $\beta < \alpha$ such that $A \in V_\beta$, and by assumption $|A| \in V_\beta$, and as $V_\beta \subseteq V_\alpha$ since $\beta < \alpha$, it follows that $|A| \in V_\alpha$.
However, the successor case is still a mystery to me, so I'm not even sure if it's true.
I would like to know which class of ordinals satisfy it, and why.
Note that for all ordinals $\alpha \colon V_{\alpha} \cap \mathrm{Ord} = \alpha$. [This follows easily by an induction on $\alpha$.]
Now $\mathcal P(\omega) \in V_{\omega+2}$ but $| \mathcal P(\omega)| \ge \omega_1$ and hence $| \mathcal P(\omega) | \not \in V_{\omega+2}$.