# Closure of Finite Union equals Union of Closures

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## Contents

## 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:

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

## Proof

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

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

We need now to show that:

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

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

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.

From Set is Subset of its Topological Closure:

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

and so $H \subseteq K$.

So from Topological Closure of Subset is Subset of Topological Closure:

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

The result follows.

$\blacksquare$

## Also see

## Sources

- 1975: W.A. Sutherland:
*Introduction to Metric and Topological Spaces*... (previous) ... (next): $3.7$: Definitions: Proposition $3.7.18$ - 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*(2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $1$: General Introduction: Closures and Interiors