Theorem 19.5 of Munkres’ Topology

Question

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.

Answer 1

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.

Answer 2

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 $\overline{\prod A_\alpha}\subseteq \prod\overline{A}_\alpha$.