For $\alpha$ a limit ordinal, show $V_{\alpha}=\bigcup_{\beta\lt\alpha}P(V_{\beta})$ For $\alpha$ a limit ordinal, I would like to show 


$V_{\alpha}=\bigcup_{\beta\lt\alpha}P(V_{\beta})$


where the $V$'s are members of the cumulative hierarchy and $P$ is the power set.
(This is a continuation of Showing equivalence of definitions for Zermelo Hierarchy, where Max ostensibly provided a solution in for a successor ordinal)
In the cumulative hierarchy, for a limit ordinal $\alpha$, $V_{\alpha}:=\bigcup_{\beta\lt\alpha}V_{\beta}$, so I would like to show 


$\bigcup_{\beta\lt\alpha}V_{\beta}=\bigcup_{\beta\lt\alpha}P(V_{\beta})$


Can I say that for any $\beta\lt\alpha$, there is a $\beta$' with $\beta\lt\beta'\lt\alpha$ such that $V_{\beta'}=P(V_{\beta})$. 
And conversely, for any such $\beta'$, there is a $\beta$ again with  $V_{\beta'}=P(V_{\beta})$, establishing inclusion in both dorections?
Thanks.

The main idea for the reverse inclusion, I think, is that $V_\beta \subset P(V_\beta)$ because $V_\beta$ is transitive. 
For the first inclusion, what you said was enough, also perhaps you should say more clearly that (with your notations) $\beta' = \beta +1 $, so that $\beta < \alpha \implies \beta' < \alpha$ (as $\alpha$ is limit)
