Closure of Union contains Union of Closures

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.

Let $\mathbb H$ be a set of subsets of $S$.

That is, let $\mathbb H \subseteq \mathcal P \left({S}\right)$ where $\mathcal P \left({S}\right)$ is the power set of $S$.

Then:
 * $\displaystyle \bigcup_{H \in \mathbb H} \operatorname{cl}\left({H}\right) \subseteq \operatorname{cl}\left({\bigcup_{H \in \mathbb H} H}\right)$

where $\operatorname{cl}\left({H}\right)$ denotes the closure of $H$ in $T$.

Proof
Let $\displaystyle K = \bigcup_{H \in \mathbb H} \operatorname{cl}\left({H}\right)$ and $\displaystyle L = \bigcup_{H \in \mathbb H} H$.

$\forall H \in \mathbb H: H \subseteq L$ so from Closure of Subset is Subset of Closure $\operatorname{cl}\left({H}\right) \subseteq \operatorname{cl}\left({L}\right)$.

It follows from the general result of Union Smallest that:
 * $K \subseteq \operatorname{cl}\left({L}\right)$