# How can I prove that there is no set containing itself without using axiom of foundation?

I've already found some similar questions in here (and other sites), but in most of the case, the use of axiom of foundation is required to complete the proof. Is there any way to prove $\not\exists x(x\in x)$, without using the axiom?

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See this comment. – Git Gud Apr 20 '14 at 12:44
Here are some lecture notes by Aczel on non-well-founded set theory. – MJD Apr 20 '14 at 13:04
The set of sets with more than one element contains itself... Right? – BlackBear Apr 20 '14 at 19:07
@BlackBear In ZF set theory, there is no such set. If Morse-Kelley set theory, there is a class of all sets with more than one element, but this class is provably not a set. I think the situation in NFU set theory is similar to MK. – MJD Apr 29 '14 at 13:30

It is consistent that the axiom of foundation fails, and there are sets of the form $x=\{x\}$. More generally there are "Anti-Foundation Axioms" which assert the existence of ill-founded sets. The more famous ones are by Boffa and by Azcel.