compact subspace of a Hausdorff space is closed

Let X be a Hausdorff space, and Y be a compactPlanetmathPlanetmath subspaceMathworldPlanetmathPlanetmath of X. We prove that XY is open, by finding for every point xXY a neighborhoodMathworldPlanetmathPlanetmath Ux disjoint from Y.

Let yY. xy, so by the definition of a Hausdorff space, there exist open neighborhoods Ux(y) of x and Vx(y) of y such that Ux(y)Vx(y)=. Clearly


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


Now for every yY there exists k1n such that yVx(yk). Since Ux(yk) and Vx(yk) are disjoint, yUx(yk), therefore neither is it in the intersectionMathworldPlanetmath


A finite intersection of open sets is open, hence Ux is a neighborhood of x disjoint from Y.

Title compact subspace of a Hausdorff space is closed
Canonical name CompactSubspaceOfAHausdorffSpaceIsClosed
Date of creation 2013-03-22 16:31:23
Last modified on 2013-03-22 16:31:23
Owner ehremo (15714)
Last modified by ehremo (15714)
Numerical id 7
Author ehremo (15714)
Entry type Proof
Classification msc 54D30