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Homeclosed set in a compact space is compact

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# 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. $\Box$

Related:

ClosedSubsetsOfACompactSetAreCompact

Major Section:

Reference

Type of Math Object:

Proof

Parent:

## Mathematics Subject Classification

54D30*no label found*

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## Comments

## mention article 4691

article 4691 called "closed subsets of a compact set are compact" deals with almost the same subject matter an should be mentioned.