Continuous Mapping from Compact Space to Hausdorff Space is Closed Mapping

Theorem

Let $T_1 = \struct {S_1, \tau_1}$ be a compact topological space.

Let $T_2 = \struct {S_2, \tau_2}$ be a $T_2$ (Hausdorff) space.

Let $f: T_1 \to T_2$ be a continuous mapping.

Then $f$ is a closed mapping.

Proof

Let $C$ be a closed subspace of $T_1$.

By Closed Subspace of Compact Space is Compact, $C$ is compact.

By Continuous Image of Compact Space is Compact, $f \sqbrk C$ is compact in $T_2$.

By Compact Subspace of Hausdorff Space is Closed, $f \sqbrk C$ is closed in $T_2$.

The result follows by definition of closed mapping.

$\blacksquare$