Can a nonempty set ever equal its Cartesian product with another set? Suppose that $S$ and $T$ are sets, with $S\neq \emptyset$. Would it be possible to have $S=S\times T = \{(s,t): s\in S, t\in T\}$? If such were the case, then we'd have
\begin{align*}
\{(s,t): s\in S, t\in T\} &= \{(s,t): s\in \{(s,t): s\in S,t\in T\}, t\in T\}\\ & = \{(s,t): s\in \{(s,t): s\in \{(s,t): s\in S,t\in T\},t\in T\}, t\in T\}\\
& = ... \text{and so on.}\\
\end{align*}
Would this make any sense? 
 A: It's not possible if we assume the usual axioms of set theory (specially the axiom of regularity). Take any $a\in A$ then $(b,t)=a$ for some $b\in A$ and some $t\in T$ then we must have $\{\{b\},\{b,t\}\}=a$ and so $b\in \{b\}\in a$, continuing this way we get an infinite chain $x_1\ni x_2\ni x_3\ni...$ which is impossible by the axiom of regularity.
A: One clear case where this happens is $\emptyset = \emptyset \times T$, for any set $T$. If $S$ is non-empty, then $S=S\times T$ is not quite a possibility, however this is a bit subtle since you need to first be very precise about what you allow as a set and what not (i.e., which model of set theory you actually use), and what precisely do you mean by the cartesian product, i.e., what is $(a,b)$. Under the usual interpretation, $S=S\times T$ for non-empty $S$ is impossible. There are however what are known as non-well-founded sets, and there such things may happen (though I'm not quite knowledgeable enough on non-well-founded sets, so this may be incorrect). 
A: If $S$ is the empty set, then $S = S \times T$ for any set $T$. See http://plato.stanford.edu/entries/nonwellfounded-set-theory/ for information on axiomatisations of set theory without the axiom of well-foundedness in which equations like $S = S \times T$ can have solutions for non-empty $S$.
A: Assuming Kuratowski's definition of ordered pairs,
$$\{a\}=\{a\}\times\{a\}\iff a=\langle a,a\rangle\iff a=\{\{a\}\}$$
which is weaker than $a=\{a\}$ and consistent with ZFC minus the axiom of foundation.
