Union of Two Compact Sets is Compact

Theorem

Let $T = \struct {S, \tau}$ be a topological spaces.

Let $H$ and $K$ be compact subsets of $T$.

Then $H \cup K$ is compact in $T$.

Proof

Let $\CC$ be an open cover of $H \cup K$.

Then $\CC$ is an open cover of both $H$ and $K$.

As $H$ and $K$ are both compact in $T$:

$H$ has a finite subcover $C_H$ of $\CC$

$K$ has a finite subcover $C_K$ of $\CC$.

Their union $C_H \cup C_K$ is a finite subcover of $\CC$ for $H \cup K$.

From Union of Finite Sets is Finite it follows that $C_H \cup C_K$ is finite.

As $\CC$ is arbitrary, it follows by definition that $H \cup K$ is compact in $T$.

Sources

• 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $5$: Compact spaces: Exercise $5.10: 3$