nLab compact subspaces of Hausdorff spaces are closed

Statement

Proposition

A compact subspace $K$ of a Hausdorff topological space $X$ is a closed subspace.

Proof

Let $x \in X \backslash K$ be any point of $X$ not contained in $K$. We need to show that there exists an open neighbourhood of $x$ in $X$ which does not intersect $K$.

By assumption that $X$ is Hausdorff, there exist for each $y \in K$ disjoint open neighbourhoods $y \subset U_y \subset X$ and $x \subset V_y \subset X$. Clearly the union of all the $U_y$ is an open cover of $K$

$$K \subset \underset{y \in K}{\cup} U_y \,.$$

Hence by assumption that $K$ is compact, there exists a finite subset $S \subset K$ of points in $K$ such that the $U_s$ for $s \in S$ still cover $K$:

$$K \subset \underset{y \in S \subset K}{\cup} U_y \,.$$

Since $S$ is finite, the intersection

$$U_x \coloneqq \underset{y \in S \subset K}{\cap} V_y$$

is still open, and by construction it is disjoint from all $U_y$ for $y \in S$, hence in particular disjoint from $K$, and it contains $x$. Hence $U_x$ is an open neighbourhood of $x$ as required.

Remarks

Remark

The statement of prop. does not hold with the Hausdorff condition replaced by weaker separation assumptions.

To see this, consider a dense subspace of a topological space. For instance the affine scheme Spec(Z) is $T_1$ but the conclusion of the proposition does not hold for the generic point.

Related entries

• closed subsets of compact spaces are compact

• closed subspaces of compact Hausdorff spaces are equivalently compact subspaces

• continuous images of compact spaces are compact

• a CW-complex is a Hausdorff space

• Hausdorff implies sober

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

References

• ProofWiki, Compact Subspace of Hausdorff Space is Closed