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Given the definition of subsets and equality of sets:

  • A $\subset$ B, if x $\epsilon$ A $\rightarrow$ x $\epsilon$ B for every set x.
  • A = B, if A $\subset$ B and B $\subset$ A

Why is it impossible to decide whether two circular sets I = {I} and J = {J} are equal.

I mean, the way is see it is that I is not an element of J, since only J is an element of J, so the two circular sets are not equal.

What's wrong in my reasoning?

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    $\begingroup$ You're assuming I≠J to get that I is not an element of J. $\endgroup$
    – user5810
    Commented Sep 8, 2010 at 11:07
  • $\begingroup$ Right, and to verify this I have to perform the same operation again and I will end up with the same problem. That makes sense. $\endgroup$
    – berater
    Commented Sep 8, 2010 at 11:24
  • $\begingroup$ Note that x={x} is explicitly forbidden by the ZF axiom of regularity (en.wikipedia.org/wiki/Axiom_of_regularity). But if decide that you are talking of another binary relation, that you still denote $\in$, but it is not the usual "is an element of" of ZF set theory, and X={X} may happen, then you also can have many such "self-singletons", why not. In particular, deciding if 2 sets are equal can't be decided iterating the procedure of checking if their elements are equal, like in your example. $\endgroup$
    – Pietro Majer
    Commented Sep 8, 2010 at 11:30
  • $\begingroup$ @berater: you could as well check that $I=J$. It would reduce to checking $J=I$ again, which makes as much sense than the converse. $\endgroup$
    – Benoît Kloeckner
    Commented Sep 8, 2010 at 11:45
  • $\begingroup$ There is some relevant discussion of non-well-founded sets and the relevant axioms of foundation, anti-foundation etc at another MO question: mathoverflow.net/questions/33282/can-we-have-aa $\endgroup$
    – Peter LeFanu Lumsdaine
    Commented Sep 8, 2010 at 13:55

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In ZF minus the axiom of foundation there is no way of proving that all "circular" sets are equal. You could take a model of set theory allowing ur-elements and replace some or all of these by circular sets.

If you accept the popular alternative to Foundation, Peter Aczel's Anti-Foundation axiom, then there is just one circular set. Effectively the AFA decrees that two sets with the same membership digraph on its transitive closure are equal.

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  • $\begingroup$ It seems that you had better replace all the urelements, since otherwise you will still have urelements remaining, which would contradict the usual understanding of ZF. $\endgroup$
    – Joel David Hamkins
    Commented Sep 8, 2010 at 12:57

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