# Paracompact 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., paracompact Hausdorff space) must also satisfy the second topological space property (i.e., normal space)

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## Statement

Any paracompact Hausdorff space (i.e., a space that is both paracompact and Hausdorff) is a normal space.

## Related facts

- Compact Hausdorff implies normal
- Paracompactness is weakly hereditary: Any closed subspace of a paracompact space is paracompact.

## Proof

**Given**: A paracompact Hausdorff space .

**To prove**: is a normal space.

**Proof**: We first prove that is a regular space, and then prove that is normal.

### Proof of regularity

**To prove**: If and is a closed set not containing , there exist open subsets such that , and is empty.

**Proof**:

- By Hausdorffness, we can define, for every , open subsets such that , and is empty.
- The open subsets , cover . In other words, .
- The sets and form an open cover of . Thus, by paracompactness of , there is a locally finite open refinement. Throwing out from this any open subset not intersecting , we still get a locally finite collection of open subsets, each contained in some , that cover .
- There exists an open set containing such that there are only finitely many members of that intersect : This follows from the definition of local finiteness.
- Let be a finite subset of that contains, for each of this finite list of members of , a point such that that member is contained in .
- Define and to be the union of all the members of . Then, , and and are disjoint: For this, note that all the members of that intersect are contained in s, which are disjoint from the corresponding s. So, is disjoint from . Finally, note that is open since it is an intersection of finitely many open subsets, and is open since it is a union of open subsets.

### Proof of normality

**To prove**: If are closed sets, there exist open sets of containing and respectively such that and are disjoint.

**Proof**:

- For every , there exist open sets containing , such that and are disjoint. This follows from regularity.
- The s form a collection of open subsets of covering . Along with , these form an open cover of . This has a locally finite open refinement. Throwing out from this any open subset not intersecting , we still get a locally finite collection of open subsets, each contained in some , that cover . Let be the union of all members of .
- For any , there exists an open subset around that does not intersect : First, there exists an open subset around intersecting only finitely many members of . Let be a finite subset of that contains, for each of this finite list of members of , a point such that that member is contained in . Then, works.
- Let be the union of all s, . Then, and are the required disjoint open subsets: This follows from the previous step.