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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.