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
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