Compact Hausdorff implies normal

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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 (i.e., compact Hausdorff space) must also satisfy the second topological space property (i.e., normal space)
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Statement

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

Related facts

Intermediate properties

Other related facts

Facts used

  1. Compactness is weakly hereditary: Any closed subset of a compact space is compact.
  2. A union of arbitrarily many open subsets is open.
  3. An intersection of finitely many open subsets is open.

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

Given: x \in X is a point and A is a closed subset of X not containing x.

To find: Disjoint open subsets containing A and x respectively.

Solution:

Step no. Assertion/construction Given data used Facts used Previous steps used Explanation
1 For every y \in A, construct disjoint open subsets V_y \ni x and U_y \ni y. In particular, the union of the U_ys contains A X is Hausdorff -- -- --
2 A is compact X is compact, A is closed in X Fact (1)
3 The U_y \cap A have a finite subcover (as an open cover of A). Thus, there is a finite set y_1,y_2,\ldots, y_r of points such that the set U = \bigcup_{i=1}^n U_{y_i} contains A. -- -- Step (2)
4 U = \bigcup_{i=1}^r U_{y_i} is an open set -- Fact (2) Step (3)
5 V = \bigcap_{i=1}^r V_{y_i} is an open set containing x -- Fact (3) Steps (1),(3) [SHOW MORE]
6 U and V are disjoint -- -- Step (1) [SHOW MORE]
7 The sets U and V are disjoint open subsets containing A and x respectively. -- -- Steps (3)--(6)

Proof of normality

Given: Disjoint closed subsets A and B of X.

To find: Disjoint open subsets U and V of X such that A \subseteq U and B \subseteq V.

Solution:

Step no. Assertion/construction Given data used Facts used Previous steps used Explanation
1 For every point x \in B, we can define disjoint open subsets U_x \supset A and V_x \ni x. In particular, V_x form an open cover of B. -- -- X is regular, by the previous half of the proof.
2 B is compact X is compact, B is closed in X Fact (1)
3 The V_x \cap B have a finite subcover, say corresponding to points x_1, x_2, \dots, x_n. Thus, the union V = \bigcup_{i=1}^n V_{x_i} contains B -- -- Step (2)
4 V = \bigcup_{i=1}^n V_{x_i} is an open subset of X -- Fact (2) Step (3)
5 U = \bigcap_{i=1}^n U_{x_i} is an open subset of X -- Fact (3) Step (3)
6 U and V are disjoint -- -- Step (1) [SHOW MORE]
7 U and V are disjoint open subsets of X containing A and B respectively. -- -- Steps (3)--(6)

References

Textbook references

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