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

Question

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?

Thank you in advance.

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

score 13 · Accepted Answer · answered Apr 20 '14 at 12:45

13

You can't.

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.

answered Apr 20 '14 at 12:45

Asaf Karagila

405,794

Uh, where exactly? – Asaf Karagila Apr 20 '14 at 12:49
1

Nevermind, I was parsing it wrongly. – Git Gud Apr 20 '14 at 12:50
Okay then. Because I've been up an ungodly number of hours, and it's very feasible that I made a typo. – Asaf Karagila Apr 20 '14 at 12:52

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

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