# Are all minimal closed sets boxes?

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.

• 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.
– Ulli
Aug 9, 2022 at 5:41
• Many thanks @Ulli. This is extremely helpful. I have included a (hopefully convincing) direct argument for reference.
– Jim
Aug 9, 2022 at 21:32

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 $$|p_1|{\times}|p_2|$$.