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closed subspaces of compact spaces are compact

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

Proposition

Let (X,τ)(X,\tau) be a compact topological space, and let YXY \subset X be a closed topological subspace. Then also YY is compact.

Proof

Let {V iY} iI\{V_i \subset Y\}_{i \in I} be an open cover of YY. We need to show that this has a finite sub-cover.

By definition of the subspace topology, there exist open subsets U iU_i of XX with

V i=U iY. V_i = U_i \cap Y \,.

By the assumption that YY is closed, the complement XYX \setminus Y is an open subset of XX, and therefore

{XYX}{U iX} iI \{ X \setminus Y \subset X\} \cup \{ U_i \subset X \}_{i \in I}

is an open cover of XX. Now by the assumption that XX is compact, this latter cover has a finite subcover, hence there exists a finite subset JIJ \subset I such that

{XYX}{U iX} iJI \{ X \setminus Y \subset X\} \cup \{ U_i \subset X \}_{i \in J \subset I}

is still an oopen cover of XX, hence in particular intersects to a finite open cover of YY. But since Y(XY)=Y \cap ( X \setminus Y ) = \empty, it follows that indeed

{V iY} iJI \{V_i \subset Y\}_{i \in J \subset I}

is a cover of YY, and in indeed a finite subcover of the original one.

Last revised on May 19, 2017 at 10:08:33. See the history of this page for a list of all contributions to it.