Are there nonempty sets that are their own binary Cartesian product?

Question

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?

Answer 1

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 $$X\times X=\{(X,X)\}=\{\{\{X\}\}\}=\{\{X\}\}=\{X\}=X.$$