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.

$b<a\implies b+1<a\implies P(V_b)=V_{b+1}\subset \cup_{c<a}V_c=V_a.$
So $\cup_{b<a}V_{b+1}\subset V_a.$ 
By transfinite induction every $V_b$ is a transitive set. If $x$ is transitive then $x\subset P(x).$ So $V_b\subset V_{b+1}.$ So $$V_a=\cup_{b<a}V_b\subset\cup_{b<a}V_{b+1}\subset V_a.$$
More generally if $S\subset a=\cup a$ then $\cup S=a\iff \cup_{b\in S}V_b=V_a.$
