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So Axiom of regularity states: every non-empty set A contains an element that is disjoint from A

I'm wondering if this is equivalent as any set is not a member of itself? If so, how do we prove it?

If not, I am wondering is knowing any set is not a member of itself sufficient to derive that a∈b and b∈a cannot be both true?

Asaf Karagila
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user1819047
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1 Answers1

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No, this is not equivalent.

It is consistent that $A\notin A$ for all $A$, but there is $A,B$ such that $A\in B\in A$. The proof is quite difficult, but we can start with a model with atoms, and an extensional relation on the atoms, then realize that relation as $\in$.

With this we can start with two atoms, and the relation will simply be $a\mathrel{E}b\mathrel{E}a$.

You can ask whether or not the axiom of regularity is equivalent to just saying that there are no infinite decreasing sequences, and without the axiom of choice this is not true either using the same constructions, only mixing in some counterexamples to the axiom of choice.

Asaf Karagila
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    Hi Asaf. Do you have a reference for this? I couldn't locate it (after a not too systematic or extensive search...) – Andrés E. Caicedo Apr 11 '14 at 16:04
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    I have the Hebrew notes of Matti Rubin. I believe this is due to Mostowski. The part about the axiom of choice is a question that came up on the site before, I told him about it and he got interested enough to sit and verify, but I'm not sure if there's any written accounts. I'll try to find something later tonight and let you know. – Asaf Karagila Apr 11 '14 at 16:06
  • Thanks! It would be nice to have an English version of those notes... Related: Was there any luck looking for references for this? – Andrés E. Caicedo Apr 11 '14 at 16:11
  • Hey Asaf, can u elaborate a little bit on how to derive a E b E a please? – user1819047 Apr 11 '14 at 16:12
  • @user1819047: You take the theorem that any extensional relation on the atoms can be realized as $\in$, and you take that given $E$. – Asaf Karagila Apr 11 '14 at 16:13
  • @Andres: I think that I couldn't find anything. I'll take another look while I'm at it. – Asaf Karagila Apr 11 '14 at 16:13
  • @Andres: Taking a quick look (since I have to go in a moment), it seems that Felgner's book "Models of ZF set theory" has a chapter about these models. Felgner attributes the consistency of $\lnot\sf Reg$ to Bernays and Specker. – Asaf Karagila Apr 11 '14 at 16:25
  • @AsafKaragila Nice. I'll check the book, every time I do I find something interesting there. – Andrés E. Caicedo Apr 11 '14 at 16:27
  • @Andres: I just checked Matti's homepage, and apparently he rewrote his notes in English when he gave the course two years ago. I'm not a big fan of his writing style (when it comes to lecture notes, anyway), and some of the terms are bad translations of Hebrew terms, which were badly translated from English before. – Asaf Karagila Apr 11 '14 at 18:30
  • Ah, great! (But I failed at finding his page... There seems to be no link from http://www.math.bgu.ac.il/directory/faculty.html, and the guess http://www.math.bgu.ac.il/~mattirubin returns a "Forbidden".) – Andrés E. Caicedo Apr 11 '14 at 19:45
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    @Andres: Without the Rubin part, just ~matti (you might have to add www. at the beginning too) Here's a direct link to the file, might save some time: http://www.math.bgu.ac.il/~matti/Axiomatic-ST-2012-09--Negation-of-Fnd-is-consistent.pdf I'm pretty sure the proof is due to Bernays, at least from the looks of it. But only an examination of his paper (and Specker's habilitation) will give a conclusive answer. – Asaf Karagila Apr 11 '14 at 20:35
  • Ah, excellent! Once again, many thanks. – Andrés E. Caicedo Apr 11 '14 at 20:53
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    @Andres: No problems. I have long contemplated of writing that construction in what I consider a readable form (I always got lost in Matti's notes, we're just too different about that). I might do that over the weekend. I'll let you know how it went. – Asaf Karagila Apr 11 '14 at 20:54