# Intersection of Closed Set with Compact Subspace is Compact

## Theorem

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

Let $H \subseteq S$ be closed in $T$.

Let $K \subseteq S$ be compact in $T$.

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

## Proof 1

Let $\tau_K$ be the subspace topology on $K$.

Let $T_K = \left({K, \tau_K}\right)$ be the topological subspace determined by $K$.

By Closed Set in Topological Subspace, $H \cap K$ is closed in $T_K$.

By Closed Subspace of Compact Space is Compact, $H \cap K$ is compact in $T_K$.

By Compact in Subspace is Compact in Topological Space, $H \cap K$ is compact in $T$.

$\blacksquare$

## Proof 2

Let $\family {U_\alpha}$ be an open cover of $H \cap K$:

- $\ds H \cap K \subseteq \bigcup_\alpha U_\alpha$

Then:

- $\ds K \subseteq \bigcup_\alpha U_\alpha \cup \paren {S \setminus H}$

Since $H$ is closed in $T$, $\paren {S \setminus H}$ is open in $T$.

Hence $\family {U_\alpha} \cup S \setminus H$ is an open cover of $K$.

We have that $K$ is compact in $T$.

It follows by definition that a finite subcover:

- $\set {U_{\alpha_1}, U_{\alpha_2}, \ldots, U_{\alpha_n}, S \setminus H}$

of $K$ exists.

Thus:

- $H \cap K \subseteq \set {U_{\alpha_1}, U_{\alpha_2}, \ldots, U_{\alpha_n} }$

and $H \cap K$ is compact in $T$.

$\blacksquare$