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 ProofThatACompactSetInAHausdorffSpaceIsClosed 2013-03-22 13:34:54 2013-03-22 13:34:54 yark (2760) yark (2760) 7 yark (2760) Proof msc 54D10 msc 54D30