# If $X \times Y$ is an Alexandrov space, both $X$ and $Y$ are Alexandrov spaces

Below is a proof (which I've struggled with for some reason) that $X \times Y$ is an Alexandrov space (arbitrary intersections of open sets are open) only if both $X$ and $Y$ are Alexandrov spaces. It uses the equivalent condition that a space $X$ is an Alexandrov space if and only if there exists a smallest open neighborhood about $x$ for every $x \in X$.

Suppose $X \times Y$ is an Alexandrov space. Let $x \in X$, $y \in Y$. We will prove that $x$ and $y$ have smallest open neighborhoods. Let $U$ be any open set in $X$ containing $x$ and let $V$ be any open set in $Y$ containing $y$. So $U \times V$ is an open set in $X \times Y$ containing $(x, y)$. Since $X \times Y$ is Alexandrov, there exists a smallest open neighborhood $M$ about $(x, y)$. Since $M$ is open, $M = \bigcup_\alpha K_\alpha \times L_\alpha$ where each $K_\alpha$ is open in $X$ and each $L_\alpha$ is open in $Y$. We know that $x$ is contained in one of these $K_\alpha$ (call it $A$) and $y$ is contained in one of these $L_\alpha$ (call it $B$). Therefore, $A \times B$ is an open set containing $(x, y)$. Since $A \times B \subseteq M$ and $M$ is the smallest open neighborhood about $(x, y)$, $M = A \times B$. Futhermore $M = A \times B \subseteq U \times V$ (since $M$ is the smallest open neighborhood), and so $A \subseteq U$ and $B \subseteq V$ (since $A$ and $B$ are nonempty). Therefore, $A$ and $B$ are the smallest open neighborhoods of $x$ and $y$ respectively. Hence $X$ and $Y$ are Alexandrov spaces.

Is this proof correct? I have a feeling that there may be quicker way...

Hint: Let $\mathcal{U}$ be any collection of open subsets of $X$, and consider the intersection of all subsets of $X\times Y$ of the form $U\times Y$ with $U\in\mathcal{U}$.
• Exactly right. And the same kind of argument works for $Y$, of course. – Cameron Buie May 9 '13 at 22:03