I'll set up a couple of (standard) definitions, then ask my question.

Definition: If $\mathbb X$ and $\mathbb Y$ are sets and $X\subseteq \mathbb X$ and $Y\subseteq \mathbb Y$, call $X{\times}Y=\{(x,y)\mid x\in X,\ y\in Y\}$ a box, and call $X$ and $Y$ the sides of the box $X{\times}Y$.

Definition: For topologies $\mathbb T_1$ and $\mathbb T_2$, the product topology $\mathbb T_1\times\mathbb T_2$ has open sets generated by boxes of open sets from $\mathbb T_1$ and $\mathbb T_2$.

Definition: Given a topology $\mathbb T$ and a point $p\in\mathbb T$, the closure of $p$, written $|p|$, is the least closed set containing $p$.

I believe it is a fact that in the product topology, the box $C_1{\times}C_2$ is closed in $\mathbb T_1\times\mathbb T_2$ if and only if its sides $C_1$ and $C_2$ are closed in $\mathbb T_1$ and $\mathbb T_2$ respectively.

My question: Given points $p_1\in\mathbb T_1$ and $p_2\in\mathbb T_2$, is the closure $|(p_1,p_2)|$ equal to the box of the closures $|p_1|{\times}|p_2|$? It's fairly clear that $|(p_1,p_2)|\subseteq |p_1|{\times}|p_2|$, but it's less immediately evident to me whether the reverse implication must also hold, so that this would be an equality.

Proof or counterexample very welcome. Thanks in advance.

  • $\begingroup$ The following more general result holds: Let $(X_i)_{i \in I}$ be a family of topological spaces, $A_i$ a subset of $X_i$ for each $i \in I$. Then $\overline{\Pi_{i \in I} A_i} = \Pi_{i \in I} \overline{A_i}$. See, for instance, Engelking, General Topology 2.3.3. $\endgroup$
    – Ulli
    Aug 9, 2022 at 5:41
  • $\begingroup$ Many thanks @Ulli. This is extremely helpful. I have included a (hopefully convincing) direct argument for reference. $\endgroup$
    – Jim
    Aug 9, 2022 at 21:32

1 Answer 1


Thanks to @Ulli for noting that this is a known result and for the precise reference.

I'll include here what seems (to me) a clean direct argument, for reference.

To prove $|(p_1,p_2)|=|p_1|{\times}|p_2|$ it would suffice to prove equality of the complements.

Say that a set $X$ avoids $x$ when $x\not\in X$, and note that $X{\times}Y$ avoids $(x,y)$ if and only if $X$ avoids $x$ and $Y$ avoids $y$.

Then the complement of $|(p_1,p_2)|$ is the union of open boxes $O_1{\times}O_2$ that avoid $(p_1,p_2)$; which is the union of open boxes $O_1{\times}O_2$ such that $O_1$ avoids $p_1$ and $O_2$ avoids $p_2$; which is the complement of $|p_1|{\times}|p_2|$.


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