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

Proof
From Closure of Union contains Union of Closures we have that:


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

We need now to show that:


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

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

As Closure is Closed, all of $\operatorname{cl}\left({H_i}\right)$ are closed.

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