Topology: union of the closure

Question

Is there any sufficient condition for the following equality?

$\overline{\bigcup_{I\in I} A_i} = \bigcup_{I\in I} \overline{A_i}$,

where $I$ is an uncountable set of index and $A_i$ are closed set for all $i\in I$.

Answer 1

Local finiteness is sufficient:

Definition A family $(A_i)_{i\in I}$ of subsets $A_i \subseteq X$ is called locally finite in $X$, iff for each $x \in X$ there is a neighbourhood $U$ of $x$ meeting only finitely many of the $A_i$.

We have:

Proposition. Let $(A_i)$ be a locally finite family of subsets of $X$. Then $$ \overline{\bigcup_i A_i} = \bigcup_i \overline{A_i} $$

Proof. As $A_i \subseteq {\bigcup_i A_i}$ for each $i$, by monotonicity of closure, we have $\overline{A_i} \subseteq \overline{\bigcup_i A_i}$ and hence $ {\bigcup_i\overline{ A_i}}\subseteq \overline{\bigcup_i A_i}$.

To prove the other inclusion we will show that $\bigcup_i \overline{A_i}$ is closed: Let $x \in X \setminus \bigcup_i \overline{A_i}$. By the local finiteness of $(A_i)$, there is an open $U\ni x$, that meets only finitely many of the $(A_i)$, say $\{A_{i_1}, \ldots, A_{i_n}\}$. Then $U\setminus \bigcup_{j=1}^n \overline{A_{i_j}}$ is an open set containing $x$ disjoint from $X \setminus \bigcup_i \overline{A_i}$. Hence $\bigcup_i \overline{A_i}$ is closed and therefore $\bigcup_i {A_i} \subseteq \bigcup_i \overline{A_i}$ implies the missing inclusion.