nLab closed subspaces of compact spaces are compact

Statement

Proposition

Let $(X,\tau)$ be a compact topological space, and let $Y \subset X$ be a closed topological subspace. Then also $Y$ is compact.

Proof

Let $\{V_i \subset Y\}_{i \in I}$ be an open cover of $Y$. We need to show that this has a finite sub-cover.

By definition of the subspace topology, there exist open subsets $U_i$ of $X$ with

$$V_i = U_i \cap Y \,.$$

By the assumption that $Y$ is closed, the complement $X \setminus Y$ is an open subset of $X$, and therefore

$$\{ X \setminus Y \subset X\} \cup \{ U_i \subset X \}_{i \in I}$$

is an open cover of $X$. Now by the assumption that $X$ is compact, this latter cover has a finite subcover, hence there exists a finite subset $J \subset I$ such that

$$\{ X \setminus Y \subset X\} \cup \{ U_i \subset X \}_{i \in J \subset I}$$

is still an open cover of $X$, hence in particular intersects to a finite open cover of $Y$. But since $Y \cap ( X \setminus Y ) = \empty$, it follows that indeed

$$\{V_i \subset Y\}_{i \in J \subset I}$$

is a cover of $Y$, and is indeed a finite subcover of the original one.

Related statements

• compact subspaces in Hausdorff spaces are separated by neighbourhoods from points

• open subspaces of compact Hausdorff spaces are locally compact

• compact subspaces of Hausdorff spaces are closed

• closed subspaces of compact Hausdorff spaces are equivalently compact subspaces

• compact Hausdorff spaces are normal

• a CW-complex is a Hausdorff space

• Hausdorff implies sober

• maps from compact spaces to Hausdorff spaces are closed and proper

• continuous images of compact spaces are compact