# Theorem 19.5 of Munkres’ Topology

Let $$\{X_\alpha\}$$ be an indexed family of spaces; let $$A_\alpha \subset X_ \alpha$$ for each $$\alpha$$. If $$\prod X_{\alpha}$$ is given either the product or the box topology, then $$\prod \bar{A}_{\alpha} = \overline{\prod A_{\alpha}}$$

It’s natural to prove $$\supseteq$$ inclusion by showing $$\prod \bar{A}_{\alpha}$$ is closed and $$\prod \bar{A}_{\alpha} \supseteq {\prod A_{\alpha}}$$. How do I show $$\prod \bar{A}_{\alpha}$$ is closed? I have tryed $$\prod (X_\alpha \setminus \overline{A_\alpha})$$, don’t known how to simplify further.

For each $$\alpha \in A$$ define $$O_\alpha = \pi_\alpha^{-1}[X_\alpha \setminus \overline{A_\alpha}]$$ which is open in the product (either topology) by continuity of the projection.

Then $$\prod_{\alpha \in A} X_\alpha \setminus \prod_{\alpha \in A} \overline{A_\alpha} = \bigcup_{\alpha \in A} O_\alpha$$ so the complement is open,hence the set is closed.

• And how to show $\prod_{\alpha \in A} X_\alpha \setminus \prod_{\alpha \in A} \overline{A_\alpha} = \bigcup_{\alpha \in A} O_\alpha$? Jan 22 at 15:07
• @user264745 if $f$ is not in the product of the closed sets it must have at least one coordinate in the complement. Just logic. Jan 22 at 15:11
• I’m sorry, I don’t get it. Can we prove it using set theory? Jan 22 at 16:58
• @user264745 It is just set theory too, de Morgan essentially. Jan 22 at 17:32
• @user264745 it’s just obvious to anyone who thinks a bit about what it means not to be in a product. The nagar ion of a universal quantifier is an existential quantifier. Jan 23 at 11:05

Let $$x\in\overline{\prod A_\alpha}$$. Considering a neighbourhood $$V$$ of $$x_\beta$$, we see that $$\pi_\beta^{-1}(V)\cap \prod A_\alpha \neq \emptyset$$, where $$\pi_\beta$$ is the projection onto the $$\beta$$-th coordinate. But then there is $$y\in \prod A_\alpha$$ with $$y_\beta\in V\cap A_\beta$$. That is, we've shown that for any neighbourhood $$V$$ of $$x_\beta$$, its intersection with $$A_\beta$$ is non-empty. But this means $$x_\beta\in\overline{A}_\beta$$. But this holds for all indices, so $$x\in \prod \overline{A}_\alpha$$. This shows $$\overline{\prod A_\alpha}\subseteq \prod\overline{A}_\alpha$$.

• I know this prove. It is given in the text. Jan 22 at 14:41