# Compact times paracompact implies paracompact

From Topospaces

This article states and proves a result of the following form: the product of two topological spaces, the first satisfying the property Compact space (?) and the second satisfying the property Paracompact space (?), is a topological space satisfying the property Paracompact space (?).

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## Contents

## Statement

### Verbal statement

The product of a compact space with a paracompact space (given the product topology), is paracompact

### Statement with symbols

Let be a compact space and a paracompact space. Then is paracompact.

## Related facts

Other results using the same proof technique:

- Compact times metacompact implies metacompact
- Compact times orthocompact implies orthocompact
- Compact times Lindelof implies Lindelof

## Facts used

- Tube lemma: If is a compact space and is a topological space. Then, given any open subset of containing for some , there exists an open subset of such that and .

## Proof

**Given**: A compact space , a paracompact space .

**To prove** If form an open cover of , there exists a locally finite open refinement of the .

**Proof**:

- For any point , there is a finite collection of that cover : Since is compact, the subspace of is also compact, so the cover by the open subsets has a finite subcover.
- Let be the union of this finite collection of open subsets . By fact (1), there exists an open subset of such that .
- The form an open cover of .
- There exists a locally finite open refinement, say of the in : This follows from the fact that is paracompact.
- We can construct a locally finite open refinement of from these:
- For each member , there exists such that . Thus, . , in turn, is a union of a finite collection of s. Thus, is the union of the intersections .
- Since the together cover , the are an open cover of that refines the s.
- Finally, we argue that is a
*locally finite*open cover: Suppose . Since is a locally finite open cover of , there exists an open subset of containing such that intersects only finitely many members of . Thus, the neighborhood intersects only finitely many s, which in turn give rise to finitely many s each. Thus, intersects only finitely many members of the open cover.