 Solution

The cumulative hierarchy is defined by transfinite recursion:Prove transitivity and nesting together by transfinite induction. The empty set is transitive. If $V_{\alpha }$ is transitive, each $x\in V_{\alpha }$ satisfies $x\subseteq V_{\alpha }$, so $V_{\alpha }\subseteq \mathcal{P}(V_{\alpha })$. If $y\in x\in \mathcal{P}(V_{\alpha })$, then $y\in V_{\alpha }\subseteq \mathcal{P}(V_{\alpha })$; thus the successor stage is transitive. At a limit, if $y\in x\in V_{\lambda }$, choose an earlier stage containing $x$ and apply its transitivity. The union is transitive and contains every earlier stage. These observations also give $V_{\alpha }\subseteq V_{\beta }$ for all $\alpha \le \beta$.

By Axiom of foundation and well-founded recursion, define the rank of a set by $\mathrm{rank}(x)=\sup \{\mathrm{rank}(y)+1:y\in x\}$. Axiom schema of replacement ensures this is an ordinal. Induction shows every element of $x$ lies in $V_{\mathrm{rank}(x)}$, soThus every set appears in the hierarchy; the argument uses foundation, not the existence of a set containing all sets.

If $\mathrm{rank}(x)=\beta$, every subset of $x$ has rank at most $\beta$, while $x$ itself belongs to $\mathcal{P}(x)$. ConsequentlyOnly successor ordinals occur. Conversely $\mathrm{rank}(V_{\beta })=\beta$, so choosing $x=V_{\beta }$ realizes every successor $\beta +1$. The exact answer is all successor ordinals, excluding zero and nonzero limit ordinals.