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...


Your proof is correct, as far as I can see, but there is an easier 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 $U\in\mathcal{U}$.

  • $\begingroup$ Exactly right. And the same kind of argument works for $Y$, of course. $\endgroup$ – Cameron Buie May 9 '13 at 22:03
  • $\begingroup$ Thanks, much simpler than my original approach! $\endgroup$ – manthanomen May 9 '13 at 22:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.