# proof that a compact set in a Hausdorff space is closed

Let $X$ be a Hausdorff space, and $C\subseteq X$ a compact subset. We are to show that $C$ is closed. We will do so, by showing that the complement $U=X\setminus C$ is open. To prove that $U$ is open, it suffices to demonstrate that, for each $x\in U$, there exists an open set $V$ with $x\in V$ and $V\subseteq U$.

Fix $x\in U$.
For each $y\in C$, using the Hausdorff assumption^{},
choose disjoint open sets ${A}_{y}$ and ${B}_{y}$ with $x\in {A}_{y}$ and $y\in {B}_{y}$.

Since every $y\in C$ is an element of ${B}_{y}$,
the collection^{} $\{{B}_{y}\mid y\in C\}$ is an open covering of $C$.
Since $C$ is compact, this open cover admits a finite subcover.
So choose ${y}_{1},\mathrm{\dots},{y}_{n}\in C$ such that
$C\subseteq {B}_{{y}_{1}}\cup \mathrm{\cdots}\cup {B}_{{y}_{n}}$.

Notice that ${A}_{{y}_{1}}\cap \mathrm{\cdots}\cap {A}_{{y}_{n}}$,
being a finite intersection^{} of open sets, is open, and contains $x$.
Call this neighborhood^{} of $x$ by the name $V$.
All we need to do is show that $V\subseteq U$.

For any point $z\in C$, we have $z\in {B}_{{y}_{1}}\cup \mathrm{\cdots}\cup {B}_{{y}_{n}}$, and therefore $z\in {B}_{{y}_{k}}$ for some $k$. Since ${A}_{{y}_{k}}$ and ${B}_{{y}_{k}}$ are disjoint, $z\notin {A}_{{y}_{k}}$, and therefore $z\notin {A}_{{y}_{1}}\cap \mathrm{\cdots}\cap {A}_{{y}_{n}}=V$. Thus $C$ is disjoint from $V$, and $V$ is contained in $U$.

Title | proof that a compact set in a Hausdorff space is closed |
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Canonical name | ProofThatACompactSetInAHausdorffSpaceIsClosed |

Date of creation | 2013-03-22 13:34:54 |

Last modified on | 2013-03-22 13:34:54 |

Owner | yark (2760) |

Last modified by | yark (2760) |

Numerical id | 7 |

Author | yark (2760) |

Entry type | Proof |

Classification | msc 54D10 |

Classification | msc 54D30 |