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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.

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    No, but I think it's intuitive it should be a true statement.2010-08-14

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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.

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    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.2010-08-15