Axiom of Foundation implies that $V=V_\infty$ Here, $V$ denotes the class of all sets, and $V_\infty$ denotes the von Neumann universe. I've heard the claim a few times that Foundation implies that every set $x$ can be found in $V_\alpha$ for some $\alpha$, but I can't find a proof of this. Does anyone know of a proof? 
 A: It suffices to show that every transitive set belongs to $V_{\infty}$, indeed if we do prove this, let $x$ be any set, then $cl(x) \subset V_\alpha$ for some $\alpha$, and then since $x\subset cl(x)$, one gets $x\in V_{\alpha+1}$ (where $cl(x)$ denotes the transitive closure of $x$) . 
So let $y$ be a transitive set and assume $y\notin V_{\infty}$. 
Then one has that $y$ is not a subset of $V_{\infty}$ indeed otherwise the application that to $x\in y$ sends the least $\alpha$ such that $x\in V_\alpha$ would be well defined, and thus by the axiom scheme of replacement would have bounded range, and thus $y\in V_{\alpha+1}$ where $\alpha$ is an upper bound of this application. 
Thus the set $z=\{x\in y\mid x\notin V_\infty\}$ is non empty, and thus by the axiom of foundation there is $x\in z$ such that $x\cap z =\emptyset$. But $x\in z$ so $x\in y$ so by transitivity of $y$, $x\subset y$, and so since $z\cap x=\emptyset$, $x\subset y\setminus z$, and thus $x\subset V_\infty$. But we've seen that this implies $x\in V_\infty$, so $x\notin z$ : a contradiction. Therefore $z$ is empty, and $y\subset V_\infty$, and so $y\in V_\infty$.
This concludes the proof.
A bit of thought will show you take the converse is true as well, i.e. if $V=V_\infty$ then the axiom of foundation holds
