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Let $ (X, \tau) $ be a topological space and $ \{A_{\alpha}: \alpha \in I \} \subset P(X)$. Verify or disprove the following: If $ \cup \overline{A_{\alpha}}$ is closed on $ X $, then $ \cup \overline {A_{\alpha}} = \overline { \cup A_{\alpha} }$.

The containment $ \cup \overline {A_{\alpha}} \subseteq \overline { \cup A_{\alpha} }$, is clear. In fact, for this containment, it is not required that $ \cup \bar{A_{\alpha}}$ be closed. I think the other contention with the condition that $ \cup \bar{A_{\alpha}}$ is closed is also true, but it's where I get stuck. Any help will be appreciated.

u0tf1s
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2 Answers2

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$\bigcup\limits_{\alpha\in I}A_\alpha\subseteq \bigcup\limits_{\alpha\in I}\overline{A_\alpha}\subseteq \overline{\bigcup\limits_{\alpha\in I}A_\alpha}$ and $\bigcup\limits_{\alpha\in I}\overline{A_\alpha}$ is closed. Therefore, $\bigcup\limits_{\alpha\in I}\overline{A_\alpha}=\overline{\bigcup\limits_{\alpha\in I}A_\alpha}$ by minimality of the closure.

  • Alternatively but very similarly, $\bigcup_\alpha \overline{A_\alpha}\subseteq \overline{\bigcup_\alpha A_\alpha}\subseteq \overline{\bigcup_\alpha\overline{A_\alpha}}$. The assumption that $\bigcup_\alpha \overline{A_\alpha}$ is closed makes the first and third sets here equal, so the second, sandwiched between them, equals the first. (Advantage of my version: No need to mention minimality. Disadvantage: Double closure notation.) – Andreas Blass Nov 06 '17 at 03:36
  • what is minimality of the closure? –  Nov 08 '17 at 21:34
  • @Isabella Minimality in the ordered set $(\mathcal C_\tau, \subseteq)$, where $\mathcal C_\tau$ is the set of closed sets of $(X,\tau)$. As in "$\overline A$ is the least closed subset of $(X,\tau)$ which contains $A$" –  Nov 08 '17 at 21:41
  • I see. OP said this $\cup \overline {A_{\alpha}} \subseteq \overline { \cup A_{\alpha}}$ was clear, is it? could you explain please? –  Nov 08 '17 at 21:47
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    @Isabella For any $\beta$, certainly $\overline{\bigcup_\alpha A_\alpha}$ is a closed set which contains $A_\beta$. By minimality of $\overline{A_\beta}$ it contains the closure. Since it contains all the $\overline{A_\beta}$-s, it contains their union. –  Nov 08 '17 at 21:55
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Let $O=X\setminus\cup\overline{A_\alpha}$ Then $O$ is an open set. If $x\in O$, then, for each $\alpha$, $O\cap A_\alpha=\emptyset$, since $O\cap\overline{A_\alpha}=\emptyset$ and $\overline{A_\alpha}\supset A_\alpha$. But then $O\cap\left(\cup A_\alpha\right)=\emptyset$ and therefore, since $O$ is a neighborhood of $x$, $x\notin\overline{\cup A_\alpha}$. So, this proves that if $x\notin\cup\overline{A_\alpha}$, then $x\notin\overline{\cup A_\alpha}$.

José Carlos Santos
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