TITLE: Are there sets $A$ and $B$ such that $A \in B$ and $B \in A$?
QUESTION [7 upvotes]: I think that the question is clear from the title:

Are there sets $A$ and $B$ such that $A \in B$ and $B \in A$?

My feeling is that the definition of any two such sets will be circular (we can't define $B$ till we define $A$ and we can't define $A$ till we define $B$), and hence will be excluded by some axiom.
On the other hand, I can't see an inherent contradiction, though I wouldn't be surprised if some form of Russell's paradox pops up, pushed behind one more set of braces.

REPLY [7 votes]: One of the Zermelo-Frankel axioms is the axiom of regularity.
The Axiom of Regularity. For every nonempty set $A$ there exists an $a\in A$ such that $a\cap A=\varnothing$.
One may use this axiom to answer your question in the negative.
Proposition. Let $A$ and $B$ be sets. Then $A\notin B$ or $B\notin A$.
Proof. Seeking a contradiction, suppose $A\in B$ and $B\in A$. Then $B\in A\cap\{A,B\}$ and $A\in B\cap\{A,B\}$. It follows that $A\cap\{A,B\}\neq\varnothing$ and $B\cap\{A,B\}\neq\varnothing$.
Now, the axiom of regularity ensures an $X\in\{A,B\}$ such that $X\cap\{A,B\}=\varnothing$. But this implies $A\cap\{A,B\}=\varnothing$ or $B\cap\{A,B\}=\varnothing$, a contradiction. Hence $A\notin B$ or $B\notin A$. $\Box$