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.

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

  • $\begingroup$ I skipped the foundation axiom because I didn't know why it was useful. This may be the reason of its existence. Please wait while I check. Thanks. $\endgroup$ – fo9bgk Aug 14 '10 at 13:24
  • $\begingroup$ Also, you should try with \left\\{A\right\\} (please note the double backslash). $\endgroup$ – fo9bgk Aug 14 '10 at 13:26
  • $\begingroup$ Thanks for the advice, I coorected my post. The foundation axiom isn't very intuitive, but it turns out very useful in some proofs. In fact, I think the best way to understand an axiom is to see how it "works" in the proofs and what are its consequences. $\endgroup$ – Cristina Aug 15 '10 at 21:28
  • 2
    $\begingroup$ One way to think of the foundation axiom is that it says sets don't go "down indefinitely"; you cannot have an infinite chain "a1 has a2 as an element, a2 has a3 as an element, a3 has a4 as an element,..." Intuitively, if you start removing the outermost brackets from the elements of your set, and continue doing this, you will eventually finish. $\endgroup$ – Arturo Magidin Aug 15 '10 at 21:31

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