# 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 |