Statements Equivalent to Axiom of Foundation

Solution 1:

First, you want to prove that if $m$ is a set and $m\subset\bigcup_\alpha V_\alpha$, then $m\in \bigcup_\alpha V_\alpha$.

Once you have that, you might be tempted to make the following argument. Let $S=V\setminus\bigcup_{\alpha} V_\alpha$. If $S$ is nonempty, then (2) says it has an $\in$-minimal element $m$. But then every element of $m$ is in $\bigcup_\alpha V_\alpha$, so by the previous paragraph, $m\in \bigcup_\alpha V_\alpha$, which is a contradiction.

Unfortunately, this argument doesn't quite work, because $S$ might not be a set (a priori, it is only a class). To get around this, use the following trick. Suppose $x\in V\setminus \bigcup_\alpha V_\alpha$, and let $T$ be the transitive closure of $\{x\}$. Let $S=T\cap (V\setminus \bigcup_\alpha V_\alpha)$ (this is a set by Separation since $T$ is a set). Now apply the argument above to this $S$.