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.
Best Answer
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.