proof that a compact set in a Hausdorff space is closed

Let X be a Hausdorff space, and CX a compact subset. We are to show that C is closed. We will do so, by showing that the complement U=XC is open. To prove that U is open, it suffices to demonstrate that, for each xU, there exists an open set V with xV and VU.

Fix xU. For each yC, using the Hausdorff assumptionPlanetmathPlanetmath, choose disjoint open sets Ay and By with xAy and yBy.

Since every yC is an element of By, the collectionMathworldPlanetmath {ByyC} is an open covering of C. Since C is compact, this open cover admits a finite subcover. So choose y1,,ynC such that CBy1Byn.

Notice that Ay1Ayn, being a finite intersectionDlmfMathworldPlanetmath of open sets, is open, and contains x. Call this neighborhoodMathworldPlanetmathPlanetmath of x by the name V. All we need to do is show that VU.

For any point zC, we have zBy1Byn, and therefore zByk for some k. Since Ayk and Byk are disjoint, zAyk, and therefore zAy1Ayn=V. Thus C is disjoint from V, and V is contained in U.

Title proof that a compact set in a Hausdorff space is closed
Canonical name ProofThatACompactSetInAHausdorffSpaceIsClosed
Date of creation 2013-03-22 13:34:54
Last modified on 2013-03-22 13:34:54
Owner yark (2760)
Last modified by yark (2760)
Numerical id 7
Author yark (2760)
Entry type Proof
Classification msc 54D10
Classification msc 54D30