How to prove that nothing is a member of itself?

I'd like to prove that $\forall x\left(x\not\in x\right)$ in the context of Morse-Kelley set theory.

Let's call $A=\left\{y:y\not\in y\right\}$. I can easily prove that $A\not\in A$. In fact, if you suppose that $A\in A$, it follows that $A\not\in A$ by definition of class comprehension. Since this is a contradiction, then the hypothesis is wrong, that is, $A\not\in A$. But I cannot do the same for an arbitrary class $k$, since I know nothing about $k$ and I cannot apply the definition of class comprehension.

Any ideas? Thanks.

• Are you sure that's a theorem of MK set theory? – Seamus Aug 14 '10 at 11:18
• No, but I think it's intuitive it should be a true statement. – fo9bgk Aug 14 '10 at 13:21

This can be proved using the foundation axiom. Suppose there exists a class $A$ such that $A \in A$, then $A$ is a set and we can consider the set $\left\{A\right\}$. The foundation axiom says that $\exists B \in \left\{A\right\}$ such that $B \cap \left\{A\right\} = \emptyset$. Since $B \in \left\{A\right\}$, $B$ must be $A$. It follows that $A \in A=B$, thus $A \in B \cap \left\{A\right\} \neq \emptyset$. And you have a contraddiction.
• Also, you should try with \left\\{A\right\\} (please note the double backslash). – fo9bgk Aug 14 '10 at 13:26