closed set in a compact space is compact

Proof. Let $A$ be a closed set in a compact space $X$. To show that $A$ is compact, we show that an arbitrary open cover has a finite subcover. For this purpose, suppose ${\{U_i\}}_{i\in I}$ be an arbitrary open cover for $A$. Since $A$ is closed, the complement of $A$, which we denote by $A^c$, is open. Hence $A^c$ and ${\{U_i\}}_{i\in I}$ 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 $A^c$ is part of $D$ or $A^c$ is not. In any case, $D\backslash \{A^c\}$ is a finite open cover for $A$, and $D\backslash \{A^c\}$ is a subcover of ${\{U_i\}}_{i\in I}$. The claim follows. $\mathrm{\square }$

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