Closure of Finite Union equals Union of Closures

Theorem
Let $T$ be a topological space.

Let $n \in \N$.

Let $\forall i \in \left[{1. . n}\right]: H_i \subseteq T$.

Then:
 * $\displaystyle \operatorname{cl}\left({\bigcup_{i=1}^n H_i}\right) = \bigcup_{i=1}^n \operatorname{cl}\left({H_i}\right)$

Proof
Let $\displaystyle K = \bigcup_{i=1}^n \operatorname{cl}\left({H_i}\right)$ and $\displaystyle H = \bigcup_{i=1}^n H_i$.

$\forall i \in \left[{1. . n}\right]: H_i \subseteq H$ so from Closure of Subset is Subset of Closure $\operatorname{cl}\left({H_i}\right) \subseteq \operatorname{cl}\left({H}\right)$.

Hence $K \subseteq \operatorname{cl}\left({H}\right)$.

This works for any set of subsets $\left\{{H_i}\right\}$.

Conversely, $K$ is the union of a finite number of closed sets.

So $K$ is itself closed, from Topology Defined by Closed Sets.

Also, $H \subseteq K$.

So from the main definition of closure, $\operatorname{cl}\left({H}\right) \subseteq K$.

The result follows.

Note
If $H$ is the union of an infinite number of sets, the result does not necessarily hold.

Let $\displaystyle H_n \subseteq \R: H_n = \left[{\frac 1 n. . 1}\right]$ for $n \ge 2$.

Then $\operatorname{cl}\left({H_n}\right) = H_n$

Also, $\displaystyle \bigcup_{n \ge 2} \operatorname{cl}\left({H_n}\right) = \bigcup_{n \ge 2} H_n = \left({0 . . 1}\right]$.

However, $\displaystyle \operatorname{cl}\left({\bigcup_{n \ge 2} H_n}\right) = \left[{0. . 1}\right]$.

So $\displaystyle \bigcup_{n \ge 2} \operatorname{cl}\left({H_n}\right) \ne \operatorname{cl}\left({\bigcup_{n \ge 2} H_n}\right)$.