TITLE: Are there nonempty sets that are their own binary Cartesian product?
QUESTION [5 upvotes]: The empty set being the trivial example of a set that is its own Cartesian power, I want to know:
Are there nonempty sets that are their own binary Cartesian product?

REPLY [4 votes]: Assuming you are using the usual Kuratowski definition of ordered pairs, this is impossible by the axiom of regularity.  Note that the rank of the ordered pair $(x,y)=\{\{x\},\{x,y\}\}$ is greater than the ranks of both $x$ and $y$ (specifically, it is $\max(\operatorname{rank}(x),\operatorname{rank}(y))+2$).  So if $X$ is a nonempty set and $x\in X$ is an element of minimal rank, every element of $X\times X$ has rank greater than the rank of $x$.  In particular, $x\not\in X\times X$, so $X\neq X\times X$.
If you don't assume the axiom of regularity, then it is consistent for there to be nonempty sets $X$ such that $X=X\times X$.  For instance, it is consistent for there to exist a set $X$ such that $X=\{X\}$, and so then $$X\times X=\{(X,X)\}=\{\{\{X\}\}\}=\{\{X\}\}=\{X\}=X.$$