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The Axiom of Regularity prevents quine atoms like $A=\{A\}$ because the intersection between $A$ and $\{A\}$ is non-empty.

However, I can't seem to find any contradiction with $A=\{\{A\}\}$ since the axiom of regularity only reaches one set containment deep.

If this set does violate the axioms, how so? If not, what is the point of the axiom of regularity? Would this be a Quine atom in ZFC?

Thanks in advance!

Theodore Astor
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    Consider the following set $$B={A,{A},{{A}}}$$ Is there an element of $B$ disjoint from it? – Alessandro Mar 28 '22 at 15:04
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    @Peter - No, you can't repair a theory by adding axioms (if you can prove an inconsistency, you can still prove it in a system with more axioms). Regularity is mostly added to make set theory itself nicer to work with (by allowing induction on the structure of sets), it's not really needed for practical applications. – Vsotvep Mar 28 '22 at 15:04
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    @Peter Regularity was not used to repair naive set theory - adding an axiom can never make something more consistent. (Instead, the fix essentially amounted to weakening full comprehension to separation/replacement.) Note that if $\mathsf{ZFC}$ itself is consistent in the first place then so is $\mathsf{ZFC}$ with regularity removed and replaced by any of a number of antifoundation axioms. Or, consider $\mathsf{NFU}$ which outright disproves regularity and is known to be consistent relative to $\mathsf{PA}$. – Noah Schweber Mar 28 '22 at 15:14
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    @Alessandro That should be an answer (unless this question turns out to be a duplicate, which wouldn't surprise me). – Noah Schweber Mar 28 '22 at 15:15

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Assume $\{\{A\}\}\in A$, then consider the following set $$B=\{A,\{A\},\{\{A\}\}\}$$ Is there an object of $B$ disjoint from it? Can you generalize this? That is, assume $A_{n-1}=\{A_n\}$ and $A_0=A$, can you generalize the definition of $B$ to prove $A_n\in A$ cannot be true?

The first use of Regularity by Zermelo only ruled out $A\in A$ and so Russell's paradox. The form we have now is to block infinite descending $\in$-chains $$A_0\ni A_1\ni\cdots\tag{*}$$since it is the consensus that only well-founded sets (sets without infinite descending $\in$-chains like $(*)$) are "all sets". I suggest you take a look at the article Believing the Axioms if you want to know more about this.

Edit Allow me to correct my bad phrasing. Disclaimer: I'm about to quote the article above but I can't say I'm 100% sure that is true. If there are historical errors in what I'm about to say, I'll gladly correct them. Anyway, as the article says

Zermelo used a weak form of the Axiom of Foundation ($A \notin A$) to block Russell's paradox in a series of lectures in the summer of 1906

As Vsotvep rightfully points out, ZF without Regularity is already enough to block Russell's paradox. The article, in fact, continues with

[...] but by 1908, he apparently felt that the form of his Separation Axiom was enough by itself, and he left the earlier axiom off his published list.

So the correct phrasing in my answer should have been "The first use of Regularity by Zermelo (in the form $A\notin A$) aimed at blocking Russell's paradox". I wasn't claiming that Regularity ruled out Russell (although I understand why it might seem that way), but that the first instance of a "Regularity"-like axiom aimed at that.

Alessandro
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  • Thank you! I see that the axiom of pairing is also super important for this too in order to be able to construct $B$ and ${A}$ as sets in the first place. – Theodore Astor Mar 28 '22 at 18:39