= Solution
Choose $v_0\in V$, using its nonemptiness. For each $u\in U$, the openness of $U\times V$ in the <product topology> supplies open $O_u\subseteq X$, $W_u\subseteq Y$ such that
$$
(u,v_0)\in O_u\times W_u\subseteq U\times V.
$$
Since $v_0\in W_u$, every point of $O_u$ belongs to $U$. Thus $U=\bigcup_{u\in U}O_u$ is open. Interchanging the factors, using a fixed $u_0\in U$, proves $V$ open. Therefore $\boxed{U\in\mathcal T_X,\quad V\in\mathcal T_Y}$. Nonemptiness matters: an empty product is open whatever the other factor is.
Back to article page
