Y is compact if and only if every open cover of Y has a finite subcover

Let X be a topological spaceMathworldPlanetmath and Y a subset of X. Then the following statements are equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath.

  1. 1.

    Y is compactPlanetmathPlanetmath as a subset of X.

  2. 2.

    Every open cover of Y (with open sets in X) has a finite subcover.

Proof. Suppose Y is compact, and {Ui}iI is an arbitrary open cover of Y, where Ui are open sets in X. Then {UiY}iI is a collectionMathworldPlanetmath of open sets in Y with union Y. Since Y is compact, there is a finite subset JI such that Y=iJ(UiY). Now Y=(iJUi)YiJUi, so {Ui}iJ is finite open cover of Y.

Conversely, suppose every open cover of Y has a finite subcover, and {Ui}iI is an arbitrary collection of open sets (in Y) with union Y. By the definition of the subspace topology, each Ui is of the form Ui=ViY for some open set Vi in X. Now UiVi, so {Vi}iI is a cover of Y by open sets in X. By assumptionPlanetmathPlanetmath, it has a finite subcover {Vi}iJ. It follows that {Ui}iJ covers Y, and Y is compact.

The above proof follows the proof given in [1].


  • 1 B.Ikenaga, Notes on Topology, August 16, 2000, available online http://www.millersv.edu/ bikenaga/topology/topnote.htmlhttp://www.millersv.edu/ bikenaga/topology/topnote.html.
Title Y is compact if and only if every open cover of Y has a finite subcover
Canonical name YIsCompactIfAndOnlyIfEveryOpenCoverOfYHasAFiniteSubcover
Date of creation 2013-03-22 13:34:07
Last modified on 2013-03-22 13:34:07
Owner mathcam (2727)
Last modified by mathcam (2727)
Numerical id 6
Author mathcam (2727)
Entry type Theorem
Classification msc 54D30