closed set in a compact space is compact

Proof. Let A be a closed setPlanetmathPlanetmath in a compact space X. To show that A is compactPlanetmathPlanetmath, we show that an arbitrary open cover has a finite subcover. For this purpose, suppose {Ui}iI be an arbitrary open cover for A. Since A is closed, the complement of A, which we denote by Ac, is open. Hence Ac and {Ui}iI together form an open cover for X. Since X is compact, this cover has a finite subcover that covers X. Let D be this subcover. Either Ac is part of D or Ac is not. In any case, D\{Ac} is a finite open cover for A, and D\{Ac} is a subcover of {Ui}iI. The claim follows.

Title closed set in a compact space is compact
Canonical name ClosedSetInACompactSpaceIsCompact
Date of creation 2013-03-22 13:34:02
Last modified on 2013-03-22 13:34:02
Owner mathcam (2727)
Last modified by mathcam (2727)
Numerical id 9
Author mathcam (2727)
Entry type Proof
Classification msc 54D30
Related topic ClosedSubsetsOfACompactSetAreCompact