Closure via sequences

Let $(X,d)$ be a metric space and $A\subseteq X$ . For $a\in X$ ,

a\in \overline{A}\quad\Longleftrightarrow\quad \exists\ (a_n)\subset A \text{ with } a_n\to a.

Remarks

Context. This is a specifically metric phenomenon (first-countability): the topological notion of closure can be detected by sequences.

Proof sketch.

If $a\in\overline{A}$ , then by ball intersections each ball $B(a;1/n)$ meets $A$ ; pick $a_n\in A\cap B(a;1/n)$ to get $a_n\to a$ .
Conversely, if $a_n\in A$ and $a_n\to a$ , then every ball around $a$ contains some $a_n$ , hence meets $A$ , so $a\in\overline{A}$ .