Closure is the smallest closed superset

Statement

Let $X$ be a topological space and let $S \subseteq X$. Then $S \subseteq \overline{S}$, the set $\overline{S}$ is closed, and each closed set $F$ of $X$ with $S \subseteq F$ satisfies $\overline{S} \subseteq F$.

Remarks

Here $\overline{S}$ is the closure of $S$. That it is closed is the statement closure is closed. This is the dual of the statement that the interior is the largest open subset. If $S \subseteq T$, then $\overline{S} \subseteq \overline{T}$.

Explanations

L1 human

Among all closed sets that contain $S$, the closure is the smallest one. For example, each closed set that contains $(0, 1)$ in $\mathbb{R}$ also contains $[0, 1]$.