nLab
continuous images 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

Idea

The image under a continuous function of a compact topological space is itself compact (cor. 1 below.)

This is a generalization of the extreme value theorem in analysis.

Statement

In fact the following more general statement holds

Lemma

Let f:(X,τ X)(Y,τ Y)f \colon (X,\tau_X) \longrightarrow (Y,\tau_Y) be a continuous function between topological spaces such that

  1. (X,τ X)(X,\tau_X) is a compact topological space;

  2. f:XYf \colon X \to Y is a surjective function.

Then also (Y,τ Y)(Y,\tau_Y) is compact.

Proof

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

By the continuity of ff the pre-images f 1(U i)f^{-1}(U_i) form an open cover {f 1(U i)X} iI\{f^{-1}(U_i) \subset X\}_{i \in I} of XX. Hence by compactness of XX, there exists a finite subset JIJ \subset I such that {f 1(U i)X} iJI\{f^{-1}(U_i) \subset X\}_{i \in J \subset I} is still an open cover of XX. Finally, by surjectivity of ff it it follows that

Y =f(X) =f(iJf 1(U i)) =iJU i \begin{aligned} Y & = f(X) \\ & = f\left( \underset{i \in J}{\cup} f^{-1}(U_i) \right) \\ & = \underset{i \in J}{\cup} U_i \end{aligned}

where we used that images of unions are unions of images.

This means that also {U iY} iJI\{U_i \subset Y\}_{i \in J \subset I} is still an open cover of YY, and in particular a finite subcover of the original cover.

Corollary

If f:XYf \colon X \longrightarrow Y is a continuous function out of a compact topological space XX which is not necessarily surjective, then we may consider its image factorization

f:Xim(f)Y f \;\colon\; X \longrightarrow im(f) \hookrightarrow Y

with im(f)im(f) regarded as a topological subspace of YY. Now by construction Xim(f)X \to \im(f) is surjective, and so lemma 1 implies that im(f)im(f) is compact.

Revised on May 19, 2017 08:40:24 by Urs Schreiber (92.218.150.85)