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

Contents

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

Lemma

(compact subspaces in Hausdorff spaces are separated by neighbourhoods from points)

Let

  1. (X,τ)(X,\tau) be a Hausdorff topological space;

  2. YXY \subset X a compact subspace.

Then for every xX\Yx \in X \backslash Y there exists

  1. an open neighbourhood U x{x}U_x \supset \{x\};

  2. an open neighbourhood U YYU_Y \supset Y

such that

  • they are still disjoint: U xU Y=U_x \cap U_Y = \emptyset.

As a direct consequence compact subspaces of Hausdorff spaces are closed.

Proof

By the assumption that (X,τ)(X,\tau) is Hausdorff, we find for every point yYy \in Y disjoint open neighbourhoods U x,y{x}U_{x,y} \supset \{x\} and U y{y}U_y \supset \{y\}. By the nature of the subspace topology of YY, the restriction of all the U yU_y to YY is an open cover of YY:

{(U yY)Y} yY. \left\{ (U_y \cap Y) \subset Y \right\}_{y \in Y} \,.

Now by the assumption that YY is compact, there exists a finite subcover, hence a finite set SYS \subset Y such that

{(U yY)Y} ySY \left\{ (U_y \cap Y) \subset Y \right\}_{y \in S \subset Y}

is still a cover.

But the finite intersection

U xsSYU x,s U_x \coloneqq \underset{s \in S \subset Y}{\cap} U_{x,s}

of the corresponding open neighbourhoods of xx is still open, and by construction it is disjoint from all the U sU_{s}, hence also from their union

U YsSYU s. U_Y \coloneqq \underset{s \in S \subset Y}{\cup} U_s \,.

Therefore U xU_x and U YU_Y are two open subsets as required.

Created on May 19, 2017 at 14:10:29. See the history of this page for a list of all contributions to it.