# 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 $y\in Y$. $x\neq y$, 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 $U_{x}^{(y)}\cap V_{x}^{(y)}=\emptyset$. Clearly

 $Y\subseteq\bigcup_{y\in Y}V_{x}^{(y)}$

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

 $Y\subseteq V_{x}^{(y_{1})}\cup\cdots\cup V_{x}^{(y_{n})}$

Now for every $y\in Y$ there exists $k\in 1...n$ such that $y\in V_{x}^{(y_{k})}$. Since $U_{x}^{(y_{k})}$ and $V_{x}^{(y_{k})}$ are disjoint, $y\not\in U_{x}^{(y_{k})}$, 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 CompactSubspaceOfAHausdorffSpaceIsClosed 2013-03-22 16:31:23 2013-03-22 16:31:23 ehremo (15714) ehremo (15714) 7 ehremo (15714) Proof msc 54D30