nLab
compact subspaces of Hausdorff spaces are closed

Context

Topology

topology (point-set topology, point-free topology)

see also differential topology, algebraic topology, functional analysis and topological homotopy theory

Introduction

Basic concepts

Universal constructions

Extra stuff, structure, properties

Examples

Basic statements

Theorems

Analysis Theorems

topological homotopy theory

Contents

Statement

Proof

Let xX\Kx \in X \backslash K be any point of XX not contained in KK. We need to show that there exists an open neighbourhood of xx in XX which does not intersect KK.

By assumption that XX is Hausdorff, there exist for each yKy \in K disjoint open neighbourhoods yU yXy \subset U_y \subset X and xV yXx \subset V_y \subset X. Clearly the union of all the U yU_y is an open cover of KK

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

Hence by assumption that KK is compact, there exists a finite subset SKS \subset K of points in KK such that the U sU_s for sSs \in S still cover KK:

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

Since SS is finite, the intersection

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

is still open, and by construction it is disjoint from all U yU_y for ySy \in S, hence in particular disjoint from KK, and it contains xx. Hence U xU_x is an open neighbourhood of xx 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 1T_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.