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.