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 $$\operatorname{cl}\left({\bigcup_{i=1}^n H_i}\right) = \bigcup_{i=1}^n \operatorname{cl}\left({H_i}\right)$$.

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

$$\forall i \in \left[{1 \,. \, . \, n}\right]: H_i \subseteq H$$ 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 $$H_n \subseteq \R: H_n = \left[{\frac 1 n \,. \, . \, 1}\right]$$ for $$n \ge 2$$.

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

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

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

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