Contents

# Contents

## 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.

## References

Last revised on April 16, 2017 at 07:42:46. See the history of this page for a list of all contributions to it.