Disproving equality of cartesian products We are to disprove the statement $X \times Y = Y \times X \iff X = Y$ but I can't think of an example where this would be false. If $X = Y$, then wouldn't the Cartesian product be the same in either direction?
 A: Clearly, if $X = Y$ then $X \times Y = Y \times X$. But the converse implication is false. 
Indeed, take $Y = \emptyset$ and let $X \neq \emptyset$. Then $X \times Y = \emptyset = Y \times X$ but $X \neq Y$.

Note that if you add the hypothesis that $X$ and $Y$ are not empty, then $X \times Y = Y \times X$ implies that $X = Y$. 
Indeed, suppose that $X \neq Y$, then $X \not\subseteq Y$ or $Y \not\subseteq X$; assume that $X \not\subseteq Y$ (the case where $Y \not \subseteq X$ is analogous), i.e. there exists $x \in X$ such that $x \notin Y$; given $y \in Y$ (it exists because $Y \neq \emptyset$), $(x,y) \in X \times Y$ but $(x,y) \notin Y \times X$ (since $x \notin Y$), hence $X \times Y \neq Y \times X$.
A: If $X=\{0\}$ and $Y=\{1,2\}$ then $(0,2) \in X \times Y$ (as $0 \in X$ and $2 \in Y$) but $(0,2) \notin Y \times X$ (because $0 \notin Y$, e.g.). Order matters a lot. In this case $X \times Y$ and $Y \times X$ have no point in common, even.
A: The important point is that a Cartesian product is a set of ordered pairs.  So if $X$ (say) contains an element $a$ which is not in $Y$, then in $X \times Y$ the element $a$ will appear in ordered pairs only as the first item of the pair, while in $Y \times X$ it will appear only as the second item of the pair.  So no pair containing $a$ in $X \times\ Y$ can ever equal any pair in $Y \times X$, even one also containing $a$.
