Compact Hausdorff implies normal
From Topospaces
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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Contents
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
Facts used
- Compactness is weakly hereditary: Any closed subset of a compact space is compact.
- A union of arbitrarily many open subsets is open.
- An intersection of finitely many open subsets is open.
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
Given: is a point and is a closed subset of not containing .
To find: Disjoint open subsets containing and respectively.
Solution:
Step no. | Assertion/construction | Given data used | Facts used | Previous steps used | Explanation |
---|---|---|---|---|---|
1 | For every , construct disjoint open subsets and . In particular, the union of the s contains | is Hausdorff | -- | -- | -- |
2 | is compact | is compact, is closed in | Fact (1) | ||
3 | The have a finite subcover (as an open cover of ). Thus, there is a finite set of points such that the set contains . | -- | -- | Step (2) | |
4 | is an open set | -- | Fact (2) | Step (3) | |
5 | is an open set containing | -- | Fact (3) | Steps (1),(3) | [SHOW MORE] |
6 | and are disjoint | -- | -- | Step (1) | [SHOW MORE] |
7 | The sets and are disjoint open subsets containing and respectively. | -- | -- | Steps (3)--(6) |
Proof of normality
Given: Disjoint closed subsets and of .
To find: Disjoint open subsets and of such that and .
Solution:
Step no. | Assertion/construction | Given data used | Facts used | Previous steps used | Explanation |
---|---|---|---|---|---|
1 | For every point , we can define disjoint open subsets and . In particular, form an open cover of . | -- | -- | is regular, by the previous half of the proof. | |
2 | is compact | is compact, is closed in | Fact (1) | ||
3 | The have a finite subcover, say corresponding to points . Thus, the union contains | -- | -- | Step (2) | |
4 | is an open subset of | -- | Fact (2) | Step (3) | |
5 | is an open subset of | -- | Fact (3) | Step (3) | |
6 | and are disjoint | -- | -- | Step (1) | [SHOW MORE] |
7 | and are disjoint open subsets of containing and 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}