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 $V$. ∎
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. |
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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 |