# 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. $\Box$

Title closed set in a compact space is compact ClosedSetInACompactSpaceIsCompact 2013-03-22 13:34:02 2013-03-22 13:34:02 mathcam (2727) mathcam (2727) 9 mathcam (2727) Proof msc 54D30 ClosedSubsetsOfACompactSetAreCompact