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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.

user264745
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2 Answers2

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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.

Henno Brandsma
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  • And how to show $\prod_{\alpha \in A} X_\alpha \setminus \prod_{\alpha \in A} \overline{A_\alpha} = \bigcup_{\alpha \in A} O_\alpha$? – user264745 Jan 22 '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. – Henno Brandsma Jan 22 '22 at 15:11
  • I’m sorry, I don’t get it. Can we prove it using set theory? – user264745 Jan 22 '22 at 16:58
  • @user264745 It is just set theory too, de Morgan essentially. – Henno Brandsma Jan 22 '22 at 17:32
  • Note that we just showed that any product of closed sets is closed in either topology. This is not true for open sets in the product topology. – Henno Brandsma Jan 22 '22 at 17:34
  • That’s nice point to keep in mind. Anyway I’m trying to show that expression by $x\in$ RHS $\Rightarrow$ $ x\in$ LHS...... I’m having difficulty showing $\subseteq$ inclusion. – user264745 Jan 22 '22 at 17:38
  • Now I’m using tuple notation to denote an element of Cartesian product. Let $(x_\alpha )\in \cup O_\alpha$. Then $(x_\alpha )\in O_\alpha$, for some $\beta \in A$. $(x_\alpha )\in \pi_{\beta}^{-1} (X_\beta \setminus \overline{A_\beta })$. So $x_\beta = \pi_{\beta}((x_\alpha ))\in X_\beta \setminus \overline{A_\beta }. Now how to show $(x_\alpha )\in$ LHS? Facing similar problem while proving other inclusion. – user264745 Jan 23 '22 at 07:34
  • Let us continue this discussion in chat. – Henno Brandsma Jan 23 '22 at 07:50
  • If you don’t mind, can I ask, how did you come up with $\prod_{\alpha \in A} X_\alpha \setminus \prod_{\alpha \in A} \overline{A_\alpha} = \bigcup_{\alpha \in A} O_\alpha$ formula? Because that formula provide “window” to write something like complement. Which I think is amazing. – user264745 Jan 23 '22 at 10:35
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    @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. – Henno Brandsma Jan 23 '22 at 11:05
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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 $\overline{\prod A_\alpha}\subseteq \prod\overline{A}_\alpha$.

Jakobian
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