# compact subspace of a Hausdorff space is closed

Let $X$ be a Hausdorff space, and $Y$ be a compact^{} subspace^{} of $X$. We prove that $X\setminus Y$ is open, by finding for every point $x\in X\setminus Y$ a neighborhood^{} ${U}_{x}$ disjoint from $Y$.

Let $y\in Y$. $x\ne y$, so by the definition of a Hausdorff space, there exist open neighborhoods ${U}_{x}^{(y)}$ of $x$ and ${V}_{x}^{(y)}$ of $y$ such that ${U}_{x}^{(y)}\cap {V}_{x}^{(y)}=\mathrm{\varnothing}$. Clearly

$$Y\subseteq \bigcup _{y\in Y}{V}_{x}^{(y)}$$ |

but since $Y$ is compact, we can select from these a finite subcover of $Y$

$$Y\subseteq {V}_{x}^{({y}_{1})}\cup \mathrm{\cdots}\cup {V}_{x}^{({y}_{n})}$$ |

Now for every $y\in Y$ there exists $k\in 1\mathrm{\dots}n$ such that $y\in {V}_{x}^{({y}_{k})}$. Since ${U}_{x}^{({y}_{k})}$ and ${V}_{x}^{({y}_{k})}$ are disjoint, $y\notin {U}_{x}^{({y}_{k})}$, therefore neither is it in the intersection^{}

$${U}_{x}=\bigcap _{j=1}^{n}{U}_{x}^{({y}_{j})}$$ |

A finite intersection of open sets is open, hence ${U}_{x}$ is a neighborhood of $x$ disjoint from $Y$.

Title | compact subspace of a Hausdorff space is closed |
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Canonical name | CompactSubspaceOfAHausdorffSpaceIsClosed |

Date of creation | 2013-03-22 16:31:23 |

Last modified on | 2013-03-22 16:31:23 |

Owner | ehremo (15714) |

Last modified by | ehremo (15714) |

Numerical id | 7 |

Author | ehremo (15714) |

Entry type | Proof |

Classification | msc 54D30 |