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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?

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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.

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  • $\begingroup$ 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. $\endgroup$
    – avir_12
    May 30, 2020 at 13:20
  • $\begingroup$ Why should $\mathcal{P}(\beta)$ be equal to $\{\beta\}$? $\endgroup$ May 30, 2020 at 14:38
  • $\begingroup$ Ouch! You're right. So my proof no longer works... $\endgroup$
    – avir_12
    May 30, 2020 at 15:58
  • $\begingroup$ @avir_12: How did you get from $\beta\in V_{\beta+1}$ to $\{\beta\}\in V_{\beta+1}$? $\endgroup$
    – Asaf Karagila
    May 30, 2020 at 17:17
  • $\begingroup$ I was wrong to use the powerset; i think that implication could be allowed by axiom of pairing. $\endgroup$
    – avir_12
    May 30, 2020 at 17:45

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