Set Closure is Smallest Closed Set/Closure Operator

Theorem
Let $S$ be a set.

Let $\operatorname{cl}: \mathcal P \left({S}\right) \to \mathcal P \left({S}\right)$ be a closure operator.

Let $T \subseteq S$.

Then $\operatorname{cl} \left({T}\right)$ is the smallest closed set (with respect to $\operatorname{cl}$) containing $T$ as a subset.

Proof
By definition, $\operatorname{cl} \left({T}\right)$ is closed.

Let $C$ be closed.

Let $T \subseteq C$.

By the definition of closure operator, $\operatorname{cl}$ is $\subseteq$-increasing.

So:
 * $\operatorname{cl} \left({T}\right) \subseteq \operatorname{cl} \left({C}\right)$

Since $C$ is closed, $\operatorname{cl} \left({C}\right) = C$.

So:
 * $\operatorname{cl} \left({T}\right) \subseteq C$

Thus $\operatorname{cl} \left({T}\right)$ is the smallest closed set containing $T$ as a subset.