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


Theorem.

Let X be a Hausdorff space, let A be a compactPlanetmathPlanetmath 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 AU and yV.

Proof.

First we use the fact that X is a Hausdorff space. Thus, for all xA there exist disjoint open sets Ux and Vx such that xUx and yVx. Then {Ux}xA is an open cover for A. Using this characterizationMathworldPlanetmath of compactness (http://planetmath.org/YIsCompactIfAndOnlyIfEveryOpenCoverOfYHasAFiniteSubcover), it follows that there exist a finite setMathworldPlanetmath A0A such that {Ux}xA0 is a finite open cover for A. Let us define

U=xA0Ux, V=xA0Vx.

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 AU and yV. To see that U and V are disjoint, suppose zU. Then zUx for some xA0. Since Ux and Vx are disjoint, z can not be in Vx, and consequently z can not be in V. ∎

The above result and proof follows [1] (Chapter 5, TheoremMathworldPlanetmath 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.
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