nLab
countably compact topological space

Context

Analysis

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

Definition

A topological space is called countably compact if every open cover consisting of a countable set of open subsets (every countable cover) admits a finite subcover, hence if there is a finite subset of the open in the original cover which still cover the space.

Examples

Properties

Proposition

A countably compact space is a limit point compact space.

Proof

Recall that a space is limit point compact if every closed discrete subspace is finite; equivalently, if every countable closed discrete subspace is finite.

Suppose XX is countably compact and AA is a countable closed discrete subspace. For each aAa \in A, choose an open neighborhood U aU_a such that U aA={a}U_a \cap A = \{a\}, and let V aV_a be the open subset U a¬AU_a \cup \neg A. Clearly we still have V aA={a}V_a \cap A = \{a\}. Also, {V a:aA}\{V_a: a \in A\} is a countable cover of XX, hence admits a finite subcover V a 1,,V a nV_{a_1}, \ldots, V_{a_n}. But then

A=( i=1 nV a i)A= i=1 n(V a iA)={a 1,,a n}A = \left(\bigcup_{i=1}^n V_{a_i} \right) \cap A = \bigcup_{i=1}^n (V_{a_i} \cap A) = \{a_1, \ldots, a_n\}

as was to be shown.

Proposition

A space that is T 1T_1 and limit point compact is countably compact.

Proof

Consider any countable open cover U 1,U 2,U_1, U_2, \ldots of XX. Put V n=U 1U nV_n = U_1 \cup \ldots \cup U_n so that V 1V 2V_1 \subseteq V_2 \subseteq \ldots, and discard repetitions, i.e., consider the maximal chain of strict inclusions

=V 0V n 1V n 2\emptyset = V_0 \subset V_{n_1} \subset V_{n_2} \subset \ldots

It suffices to show this maximal chain is finite. Rename it as =W 0W 1W 2\emptyset = W_0 \subset W_1 \subset W_2 \subset \ldots.

For each W nW_n occurring in this chain (n1n \geq 1), pick a point x nW nW n1x_n \in W_n \setminus W_{n-1}. Observe that if m<nm \lt n, then x nW mx_n \notin W_m.

Since XX is T 1T_1 (points are closed), the set W m¬{x 1,,x m1}W_m \cap \neg \{x_1, \ldots, x_{m-1}\} is an open neighborhood of x mx_m that does not contain x nx_n whenever n>mn \gt m, and does not contain x nx_n whenever n<mn \lt m. Thus every point x mx_m is open relative to A={x 1,x 2,}A = \{x_1, x_2, \ldots\}, i.e., AA is a discrete subspace.

Finally, any point xAx \notin A belongs to some W nW_n, and then W n¬{x 1,x n}W_n \cap \neg \{x_1, \ldots x_n\} is an open neighborhood of xx that doesn’t intersect AA. Thus AA is a closed discrete subspace, and is finite by limit point compactness. Therefore the maximal chain consisting of the sets W nW_n is finite, as was to be shown.

Last revised on July 28, 2018 at 11:16:44. See the history of this page for a list of all contributions to it.