point and a compact set in a Hausdorff space have disjoint open neighborhoods.

Theorem.

Let $X$ be a Hausdorff space, let $A$ be a compact non-empty set in $X$, and let $y$ a point in the complement of $A$. Then there exist disjoint open sets $U$ and $V$ in $X$ such that $A\mathrm{\subset }U$ and $y\mathrm{\in }V$.

Proof.

First we use the fact that $X$ is a Hausdorff space. Thus, for all $x\in A$ there exist disjoint open sets $U_x$ and $V_x$ such that $x\in U_x$ and $y\in V_x$. Then ${\{U_x\}}_{x\in A}$ is an open cover for $A$. Using this characterization of compactness (http://planetmath.org/YIsCompactIfAndOnlyIfEveryOpenCoverOfYHasAFiniteSubcover), it follows that there exist a finite set $A_0\subset A$ such that ${\{U_x\}}_{x\in A_0}$ is a finite open cover for $A$. Let us define

$U={\displaystyle \bigcup _{x\in A_0}}{U}_x,$ $\mathrm{\hspace{0.17em}}V={\displaystyle \bigcap _{x\in A_0}}{V}_x.$

Next we show that these sets satisfy the given conditions for $U$ and $V$. First, it is clear that $U$ and $V$ are open. We also have that $A\subset U$ and $y\in V$. To see that $U$ and $V$ are disjoint, suppose $z\in U$. Then $z\in U_x$ for some $x\in A_0$. Since $U_x$ and $V_x$ are disjoint, $z$ can not be in $V_x$, and consequently $z$ can not be in $V$. ∎

The above result and proof follows [1] (Chapter 5, Theorem 7) or [2] (page 27).

Title	point and a compact set in a Hausdorff space have disjoint open neighborhoods.
Canonical name	PointAndACompactSetInAHausdorffSpaceHaveDisjointOpenNeighborhoods
Date of creation	2013-03-22 13:34:27
Last modified on	2013-03-22 13:34:27
Owner	drini (3)
Last modified by	drini (3)
Numerical id	13
Author	drini (3)
Entry type	Theorem
Classification	msc 54D30
Classification	msc 54D10