Compact Hausdorff implies normal: Difference between revisions

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Suppose <math>A</math> and <math>B</math> are disjoint closed subsets of <math>X</math>. Then, for every point <math>x \in B</math>, we can define disjoint open subsets <math>U_x \supset A</math> and <math>V_x \supset x</math>. The open sets <math>V_x</math> yield an open cover of <math>B</math>, which is a closed, hence compact, subset of <math>X</math>, so there is a finite subcover <math>x_1,x_2,\ldots,x_n</math>. The intersection of the <math>U_{x_i}</math>s and the union of the <math>V_{x_i}</math>s are disjoint open subsets containing <math>A</math> and <math>B</math>.
Suppose <math>A</math> and <math>B</math> are disjoint closed subsets of <math>X</math>. Then, for every point <math>x \in B</math>, we can define disjoint open subsets <math>U_x \supset A</math> and <math>V_x \supset x</math>. The open sets <math>V_x</math> yield an open cover of <math>B</math>, which is a closed, hence compact, subset of <math>X</math>, so there is a finite subcover <math>x_1,x_2,\ldots,x_n</math>. The intersection of the <math>U_{x_i}</math>s and the union of the <math>V_{x_i}</math>s are disjoint open subsets containing <math>A</math> and <math>B</math>.
==References==
===Textbook references===
* {{booklink|Munkres}}, Page 202 (Theorem 32.3)
* {{booklink|SingerThorpe}}, Page 29 (Remark)

Revision as of 20:27, 21 April 2008

This article gives the statement and possibly, proof, of an implication relation between two topological space properties. That is, it states that every topological space satisfying the first topological space property must also satisfy the second topological space property
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Statement

Property-theoretic statement

The property of topological spaces of being compact Hausdorff implies, or is stronger than, the property of being normal.

Verbal statement

Any compact Hausdorff space (a topological space that is both a compact and Hausdorff) is normal.

Related facts

Intermediate properties

Other related facts

Proof

Suppose X is a compact Hausdorff space. We need to show that X is normal. We will proceed in two steps: we will first show that X is a regular space, and then show that X is normal.

Proof of regularity

Suppose xX is a point and A is a closed subset of X not containing x. Our goal is to exhibit disjoint open subsets containing A and x respectively.

Let's do this. For every yA, use the fact that X is a Hausdorff space to construct disjoint open subsets Vyx and Uyy.

Now, using the fact that X is compact, A is closed in X, and a closed subset of a compact space is compact, we deduce that A is compact, so the UyA have a finite subcover (as an open cover of A. Thus, there is a finite set y1,y2,,yr of points such that the union of Uyi is an open subset of X containing A.

Then, U=i=1nUyi and V=i=1nVyi are disjoint open subsets containing A and x respectively.

Proof of normality

The process of jacking up from regularity to normality is completely analogous to the process of jacking up from Hausdorffness to regularity. Let's describe this process.

Suppose A and B are disjoint closed subsets of X. Then, for every point xB, we can define disjoint open subsets UxA and Vxx. The open sets Vx yield an open cover of B, which is a closed, hence compact, subset of X, so there is a finite subcover x1,x2,,xn. The intersection of the Uxis and the union of the Vxis are disjoint open subsets containing A and B.

References

Textbook references

  • Topology (2nd edition) by James R. MunkresMore info, Page 202 (Theorem 32.3)
  • Lecture Notes on Elementary Topology and Geometry (Undergraduate Texts in Mathematics) by I. M. Singer and J. A. ThorpeMore info, Page 29 (Remark)