# Axiom of power set in von Neumann hierarchy of sets.

My question is this: which stages of von Neumann universe does satisfy axiom of power set? I recall that von Neumann universe is defined by transfinite recursion: $$V_0=\emptyset; V_{\alpha+1}=\mathcal{P}(V_{\alpha}); V_{\lambda}=\bigcup_{\gamma < \lambda}V_{\gamma}$$ if $$\lambda$$ is a limit ordinal.
My claim is that stage $$V_{\alpha}$$ satisfies axiom of power set iff $$\alpha$$ is a limit ordinal. If we have $$A \in V_{\alpha}$$ (with $$\alpha$$ limit), so by definition, exists $$\gamma < \alpha$$ such that $$A \in V_{\gamma}$$. I have to prove that also $$\mathcal{P}(A) \in V_{\alpha}$$. How could I do this?

Notice that if $$A \in V_\gamma$$, then $$\mathcal{P}(A) \subseteq V_\gamma$$ and hence $$\mathcal{P}(A) \in V_{\gamma +1}$$. Since $$\alpha$$ is limit, $$\gamma +1<\alpha$$ and you're done.
• Thanks! For the converse, I thought of this counterexample: if $\alpha=\beta+1$ successor, then $\beta \subseteq V_{\beta} \Rightarrow \beta \in V_{\beta+1}$. So, $\mathcal{P}(\beta)=\{\beta\} \in V_{\beta+1}=\mathcal{P}(V_{\beta})$ from which $\{\beta\} \subseteq V_{\beta} \Rightarrow \beta \in V_{\beta}$, absurd. May 30, 2020 at 13:20
• Why should $\mathcal{P}(\beta)$ be equal to $\{\beta\}$? May 30, 2020 at 14:38
• @avir_12: How did you get from $\beta\in V_{\beta+1}$ to $\{\beta\}\in V_{\beta+1}$? May 30, 2020 at 17:17