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I thought that the regularity axiom states that there are no sets that contais themselfs. That means that it cannot happend that $A\in A$, being $A$ a set. But actually the regularity axiom states the following:

$$ \forall A (A\ne \emptyset \longrightarrow \exists B(B\in A \land B\cap A = \emptyset)) $$

That means that any non-empty set A has at least one element that is disjoint from A. This axiom, along with the axiom of parity, implies the statement that I made at the beginning.

But I was wondering if, considering the rest of the ZFC axioms, these two statements (the actual axiom of regularity and what I thought it was) are actually equivalent. I wasn't able to prove it, nor could I find a counterexample.

Gaza
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  • Interesting question. The regularity axiom states exactly that $\in$ is a well-founded collection and based on that it can be shown that also its transitive closure $\bar{\in}$ is well-founded. I don't know but have my doubts on the question whether this can be proved purely based on "sets do not contain itself as element". Just like you I hope for an answer. – drhab Jul 06 '23 at 08:34
  • This is definitely a duplicate. Hang on. – Asaf Karagila Jul 06 '23 at 08:56
  • Okay, I think what I had in mind was https://math.stackexchange.com/questions/3995586/how-much-damage-is-done-if-the-axiom-of-regularity-is-replaced-with-no-set-can which is very closely related, but not quite-right-a duplicate, and in any case it does not have an answer either. – Asaf Karagila Jul 06 '23 at 09:11
  • Ah, but https://math.stackexchange.com/questions/4519567/does-the-nonexistence-of-in-cycles-imply-the-nonexistence-of-an-infinite-desc/ is an even stronger statement. – Asaf Karagila Jul 06 '23 at 09:15

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