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 $y_{1},\ldots,y_{n}\in C$ such that $C\subseteq B_{y_{1}}\cup\cdots\cup B_{y_{n}}$ .

Notice that $A_{y_{1}}\cap\cdots\cap A_{y_{n}}$ , 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 $z\in B_{y_{1}}\cup\cdots\cup B_{y_{n}}$ , 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 $z\notin A_{y_{1}}\cap\cdots\cap A_{y_{n}}=V$ . Thus $C$ is disjoint from $V$ , and $V$ is contained in $U$ .

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