In Zermelo-Fraenkel set theory do there exist sets $a,b$ such that $a \in b$ and $b \in a$. I think that no such sets exist but I am not sure how to prove why this is the case.
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I feel like it's going wrong at the axiom of regularity: "Every non-empty set $x$ contains a member $y$ such that $x$ and $y$ are disjoint sets." – Mar 24 '17 at 08:30
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So do you mean that a would contain a subset, namely 'b', but then a and b could not be disjoint? – Prince M Mar 24 '17 at 08:32
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This is not possible due to the axiom of regularity. Define $c = \{a,b\}$ and suppose $a\in b$ and $b\in a$. Then both $a$ and $c$ contain $b$ and thus the two are not disjoint. Similarly $a\in c\cap b$. Therefore $c$ does not contain a member that is disjoint with $c$, which means the axiom of regularity doesn't hold.
Marc
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