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[parent] closed set in a compact space is compact (Proof)

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$



"closed set in a compact space is compact" is owned by mathcam. [ owner history (1) ]
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See Also: closed subsets of a compact set are compact


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Cross-references: cover, open, complement, closed, subcover, finite, open cover, compact, closed set, proof

This is version 6 of closed set in a compact space is compact, born on 2003-04-11, modified 2003-06-19.
Object id is 4177, canonical name is AClosedSetInACompactSpaceIsCompact.
Accessed 2998 times total.

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AMS MSC: 54D30 (General topology :: Fairly general properties :: Compactness)
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mention article 4691 by Linas on 2007-03-04 10:25:56
article 4691 called "closed subsets of a compact set are compact" deals with almost the same subject matter an should be mentioned.
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