# When is the closure of an intersection equal to the intersection of closures?

We know $\overline{\bigcap A_{\alpha}}\subseteq\bigcap\overline{A}_{\alpha}$, but when is the reverse inclusion true? Can you give some properties of the underlying space that would guarantee this?

This extends the answers of hardmath and Brian M. Scott, but completely answers the question.

Spaces satisfying the (seemingly) weaker condition that $\overline{ A \cap B } = \overline{A} \cap \overline{B}$ for all $A , B \subseteq X$ are discrete.

If $A \subseteq X$ is a non-closed set, pick $x \in \overline{A} \setminus A$. Note, then, that which is absurd! Therefore all subsets are closed.

As hardmath noted, all discrete spaces satisfy the stronger condition in the OP, and so we have an equivalence of all three notions.

Given a topological space $X$, the following are equivalent:

1. $\overline{ \bigcap_{i \in I} A_i } = \bigcap_{i \in I} \overline{A_i}$ for all families $\{ A_i \}_{i \in I}$ of subsets of $X$.
2. $\overline{ A \cap B } = \overline{A} \cap \overline{B}$ for all $A , B \subseteq X$.
3. $X$ is discrete.