# 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 $A\subset U$ and $y\in V$.

###### 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

 $\displaystyle U=\bigcup_{x\in A_{0}}U_{x},$ $\displaystyle\,\,\,\,\,V=\bigcap_{x\in A_{0}}V_{x}.$

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).

## References

• 1 J.L. Kelley, General Topology, D. van Nostrand Company, Inc., 1955.
• 2 I.M. Singer, J.A.Thorpe, Lecture Notes on Elementary Topology and Geometry, Springer-Verlag, 1967.
Title point and a compact set in a Hausdorff space have disjoint open neighborhoods. PointAndACompactSetInAHausdorffSpaceHaveDisjointOpenNeighborhoods 2013-03-22 13:34:27 2013-03-22 13:34:27 drini (3) drini (3) 13 drini (3) Theorem msc 54D30 msc 54D10