# Closure of Finite Union equals Union of Closures

## Theorem

Let $T$ be a topological space.

Let $n \in \N$.

Let:

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

Then:

$\ds \bigcup_{i \mathop = 1}^n \map \cl {H_i} = \map \cl {\bigcup_{i \mathop = 1}^n H_i}$

where $\map \cl {H_i}$ denotes the closure of $H_i$.

## Proof

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

$\ds \bigcup_{i \mathop = 1}^n \map \cl {H_i} \subseteq \map \cl {\bigcup_{i \mathop = 1}^n H_i}$

We need now to show that:

$\ds \bigcup_{i \mathop = 1}^n \map \cl {H_i} \supseteq \map \cl {\bigcup_{i \mathop = 1}^n H_i}$

Let $\ds K = \bigcup_{i \mathop = 1}^n \map \cl {H_i}$ and $\ds H = \bigcup_{i \mathop = 1}^n H_i$.

From Topological Closure is Closed, all of $\map \cl {H_i}$ 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.

$\forall i \in \closedint 1 n: H_i \subseteq \map \cl {H_i}$

and so $H \subseteq K$.

$\map \cl H \subseteq \map \cl K$

The result follows.

$\blacksquare$