Compact Hausdorff implies normal: Difference between revisions
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Suppose <math>x \in X</math> is a point and <math>A</math> is a [[closed subset]] of <math>X</math> not containing <math>x</math>. Our goal is to exhibit disjoint open subsets containing <math>A</math> and <math>x</math> respectively. | Suppose <math>x \in X</math> is a point and <math>A</math> is a [[closed subset]] of <math>X</math> not containing <math>x</math>. Our goal is to exhibit disjoint open subsets containing <math>A</math> and <math>x</math> respectively. | ||
Let's do this. For every <math>y \in A</math>, use the fact that <math>X</math> is a [[Hausdorff space]] to | Let's do this. For every <math>y \in A</math>, use the fact that <math>X</math> is a [[Hausdorff space]] to construct disjoint open subsets <math>V_y \ni x</math> and <math>U_y \ni y</math>. | ||
Now, using the fact that <math>X</math> is compact, <math>A</math> is closed in <math>X</math>, and [[compactness is weakly hereditary|a closed subset of a compact space is compact]], we deduce that <math>A</math> is compact, so the <math>U_y \cap A</math> have a finite subcover (as an open cover of <math>A</math>. Thus, there is a finite set <math>y_1,y_2,\ldots, y_r</math> of points such that the union of <math>U_{y_i}</math> is an open subset of <math>X</math> containing <math>A</math>. | |||
Then, <math>U = \bigcup_{i=1}^n U_{y_i}</math> and <math>V = \bigcap_{i=1}^n V_{y_i}</math> are disjoint open subsets containing <math>A</math> and <math>x</math> 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 <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>. | |||
Revision as of 20:51, 23 January 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
View all topological space property implications | View all topological space property non-implications
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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
- Paracompact Hausdorff space: Further information: paracompact Hausdorff implies normal
- any closed subset of a compact space is compact
- any compact subset of a Hausdorff space is closed: The proof of this uses a very similar argument.
- Any locally compact Hausdorff space is completely regular
Proof
Suppose is a compact Hausdorff space. We need to show that is normal. We will proceed in two steps: we will first show that is a regular space, and then show that is normal.
Proof of regularity
Suppose is a point and is a closed subset of not containing . Our goal is to exhibit disjoint open subsets containing and respectively.
Let's do this. For every , use the fact that is a Hausdorff space to construct disjoint open subsets and .
Now, using the fact that is compact, is closed in , and a closed subset of a compact space is compact, we deduce that is compact, so the have a finite subcover (as an open cover of . Thus, there is a finite set of points such that the union of is an open subset of containing .
Then, and are disjoint open subsets containing and 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 and are disjoint closed subsets of . Then, for every point , we can define disjoint open subsets and . The open sets yield an open cover of , which is a closed, hence compact, subset of , so there is a finite subcover . The intersection of the s and the union of the s are disjoint open subsets containing and .