nLab compact Hausdorff spaces are normal

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

In fact a stronger statement holds: paracompact Hausdorff spaces are normal.

To prove this, consider the following lemma:

Lemma

(separation by neighbourhoods of points from compact subspaces in Hausdorff spaces)

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.
Proof

of lemma

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}{\bigcap} 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 in particular from their union

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

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

Proof

of prop.

First we claim that (X,τ)(X,\tau) is regular. To show this, we need to find for each point xXx \in X and each disjoint closed subset YXY \in X dijoint open neighbourhoods U x{x}U_x \supset \{x\} and U YYU_Y \supset Y. But since closed subspaces of compact spaces are compact, the subset YY is in fact compact, and hence this is in fact the statement of lemma .

Next to show that (X,τ)(X,\tau) is indeed normal, we apply the idea of the proof of lemma once more:

Let Y 1,Y 2XY_1, Y_2 \subset X be two disjoint closed subspaces. By the previous statement then for every point y 1Yy_1 \in Y we find disjoint open neighbourhoods U y 1{y 1}U_{y_1} \subset \{y_1\} and U Y 2,y 1Y 2U_{Y_2,y_1} \supset Y_2. The union of the U y 1U_{y_1} is a cover of Y 1Y_1, and by compactness of Y 1Y_1 there is a finite subset SYS \subset Y such that

U Y 1sSY 1U y 1 U_{Y_1} \coloneqq \underset{s \in S \subset Y_1}{\bigcup} U_{y_1}

is an open neighbourhood of Y 1Y_1 and

U Y 2sSYU Y 2,s U_{Y_2} \coloneqq \underset{s \in S \subset Y}{\bigcap} U_{Y_2,s}

is an open neighbourhood of Y 2Y_2, and both are disjoint.

References

Last revised on March 8, 2020 at 22:52:54. See the history of this page for a list of all contributions to it.