We know that let $\{A_{i}\}_{i\in\mathbb{N}}$ a sequence of sets in a topological space with euclidean topology
\begin{equation} \overline{\bigcup_{i=1}^{+\infty}A_{i}}=\bigcup_{i=1}^{+\infty}\bar{A_{i}}\cup \bigcap_{i=1}^{+\infty}\overline{\bigcup_{j=0}^{+\infty}A_{i+j}}. \end{equation}
Do exists conditions such that
\begin{equation} \overline{\bigcup_{i=1}^{+\infty}A_{i}}=\bigcup_{i=1}^{+\infty}\bar{A_{i}}. \end{equation} Where $\bar{A_{i}}$ represents the closure of $A_{i}$
Thanks!
One condition is quite natural: We have that $A_n$ is a so-called closure preserving family (which implies your identity), when $\{A_n: n=1,2,3\}$ is locally finite. This means that for every $x \in X$, there exists an open neighbourhoods $O_x$ of $x$ such that $\{n : O_x \cap A_n \neq \emptyset \}$ is finite.