# Closure of Finite Union equals Union of Closures

## Theorem

Let $T$ be a topological space.

Let $n \in \N$.

Let:

$\forall i \in \left\{ {1, 2, \ldots, n}\right\}: H_i \subseteq T$

Then:

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

## Proof

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

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

We need now to show that:

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

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

From Topological 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.

$\forall i \in \left[{1 \,.\,.\, n}\right]: H_i \subseteq \operatorname{cl}\left({H_i}\right)$

and so $H \subseteq K$.

So from Topological Closure of Subset is Subset of Topological Closure, $\operatorname{cl}\left({H}\right) \subseteq \operatorname{cl}\left({K}\right)$.

The result follows.

$\blacksquare$