Intersection of Closed Set with Compact Subspace is Compact/Proof 1

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

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 $T$.

$\blacksquare$