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So, I'm currently taking a set theory course. I understand the axiom of regularity or foundation, and why it was added, but I can't help but get the feeling that it's too strong. I know it rules out stuff like $x\in x$ or $x\in y$ and $y\in x$. My question is, are there axioms that are more intuitive, along the lines of ruling out these cases, which are equivalent to the axiom of regularity, under the other axioms of ZFC?

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Which statement of regularity are you using? – Chris Eagle Jan 24 '13 at 18:43
I quite like V = WF - i.e. the cumulative hierarchy captures every set. – Kris Jan 24 '13 at 18:49
I'm not sure why you would need to rule out $x\in x$, etc. It never seems to come up everyday mathematics. And you really don't need to do so to avoid the contradictions of naive set theory, e.g. Russell's Paradox. – Dan Christensen Jan 25 '13 at 18:56

You can use $\textbf{There exists a rank function}$. Namely there is a formula $\varphi(x,\alpha)$ which is true if and only if $\alpha$ is an ordinal, and $\alpha=\sup\{\beta+1\mid\exists y\in x\varphi(y,\beta)\}$.

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There is no infinite sequence of sets $(x_1,x_2,\ldots)$ such that $x_1\ni x_2\ni x_3\ni\ldots$ The equivalence to the axiom of regularity requires the axiom of choice.

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While it is true in ZFC, well-foundedness is stronger than just no decreasing sequence without the axiom of choice. That been said, I am not sure that in the case of $\in$ it is true; i.e. it might be provable from the above that $\in$ is well-founded even in ZF. – Asaf Karagila Jan 24 '13 at 19:13
@Asaf You need something like DC to iteratively get $in$-smaller-and-smaller elements in case regularity is violated. – Michael Greinecker Jan 24 '13 at 19:16
I think that DC is equivalent to the equivalency between the forms of well-foundedness (it certainly implies it). But it is consistent that there are non-well founded relations without decreasing chains. My point was that it might be provable that if such set exists then there is one with an actual decreasing chain, in which case ZF-Reg is enough to prove the equivalence in the case of the $\in$ relation. – Asaf Karagila Jan 24 '13 at 19:19
@Asaf I'm not sure I get your point. Is something wrong in my answer? – Michael Greinecker Jan 24 '13 at 19:24
No, not at all. The question also specified ZFC-Reg as the theory. But I wanted to point out that well-foundedness is not equivalent to the lack of decreasing sequences without some form of choice. I also added that in the particular case of $\in$ it might be provable after all, but I don't know for sure. :-) – Asaf Karagila Jan 24 '13 at 19:27

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