TITLE: If $ \cup \bar{A_{\alpha}}$ is closed on $ X $, then $ \cup \overline {A_{\alpha}} = \overline { \cup A_{\alpha} }$
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: 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 $x\notin\overline{\cup A_\alpha}$.