closed set in a compact space is compactAClosedSetInACompactSpaceIsCompact2003-04-11 14:49:382003-06-19 06:15:48Proofproperties of compact spaces0% this is the default PlanetMath preamble. as your knowledge
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% define commands here\emph{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. $\Box$