# The von Neumann universe is equal to the class of well-founded sets

The class of well-founded sets $WF$ is defined as $x \in WF$ if there exists an ordinal $\alpha$ such that $x \in V_\alpha$. The von Neumann universe is defined as $V = \bigcup_{\alpha \in \mathbf{Ord}} V_\alpha$. I want to show that $V = WF$. Here is a proof I came up with.

Take arbitrary $x \in V$. Hence, $x \in \bigcup_{\alpha \in \mathbf{Ord}} V_\alpha$, so there exists some $\alpha$ such that $x \in V_\alpha$. Hence, $x \in WF$. The reverse direction is similarly trivial.

Now I suspect that this proof is wrong (where?) since the proof my professor gave (which I fail to understand) is more complicated, and involves assuming that $V$ satisfies the axiom of foundation. Here is his proof.

Suppose $V \models \mathrm{Foundation}$. Let $x \in V$ and suppose $x \notin WF$. First without loss of generality assume that $x$ is transitive. (Replace it with $\operatorname{trcl} x$ if necessary.) Let $y \in x$ be of minimal rank such that $y \notin WF$. Then every element of $y$ is in $WF$ since they are of lower rank. Hence $y \subseteq WF$. But then $y \in WF$ as well, since if $\alpha$ is such that every element of $y$ is in $V_\alpha$, then $y \in V_{\alpha + 1}$. This contradicts that $y \notin WF$ and hence that $x \notin WF$.

First, the reverse direction of the inclusion isn't shown. Is that because the obvious proof for it applies? (Specifically suppose $x \in WF$. Then there exists $\alpha$ such that $x \in V_\alpha$, so of course $x$ is in the union of all the $V_\alpha$.)

Second, where is the assumption that $V \models \mathrm{Foundation}$ used?

Your proof is completely fine for the definitions you have stated. Your professor is proving a different theorem, with different definitions. Specifically, it appears that your professor is defining $V$ to be the class of all sets, and is proving that if $V$ satisfies Foundation, then it is equal to $WF$.
With this definition, the reverse inclusion is totally trivial (by definition, $V$ contains all sets). The assumption that $V$ satisfies Foundation is being used to refer to "rank", and in particular to find $y\in x$ of minimal rank. You can't define the rank of an arbitrary set without using Foundation.