# 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

Let be a compact space and a paracompact space. Then , the Cartesian product endowed with the product topology, 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: Suppose 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 . form an open cover of .

**To prove**: There exists a locally finite open refinement of the s, i.e., an open cover of such that:

- It is locally finite: For any point , there exists an open set containing that intersects only finitely many of the s.
- It refines : Every is contained in one of the s.

**Proof**:

Step no. | Assertion/construction | Facts used | Given data used | Previous steps used | Explanation |
---|---|---|---|---|---|

1 | For any point , there is a finite collection of that cover | is compact | Since is compact, the subspace of is also compact, so the cover by the open subsets has a finite subcover. | ||

2 | For any point , let be the union of this finite collection of open subsets as obtained in Step (1). There exists an open subset of such that and | Fact (1) | is compact | Step (1) | Follows from Fact (1), setting the of Fact (1) to be . |

3 | The open subsets obtained in Step (2) form an open cover of . | Step (2) | By Step (2), , hence . Since , and , we get . | ||

4 | There exists a locally finite open refinement of the in | is paracompact | Step (3) | Step-given combination direct. | |

5 | For each , is a union of finitely many intersections , all of which are open subsets | Steps (1) (2), (4) | Since s refine s (Step (4)), there exists such that . In turn, by the definition of (Step (2)), we have , which in turn is a union of finitely many s (Step (1)). Thus, is contained in a union of finitely many s, and hence, is the union of its intersection with those s. Since are all open, the intersections are all open. | ||

6 | The open subsets of the form of Step (5) form an open cover of that refines the s (note that not every combination of and is included -- only the finitely many s needed as in Step (5)). We will index this open cover by indexing set , and call it , where . In particular, if , then , and for any with . | Steps (4), (5) | cover , so cover . By Step (5), is the union of the , so the latter also form an open cover of . | ||

7 | The open cover of Step (6) is a locally finite open cover. In other words, for any , there is an open subset such that intersects only finite many s. | Steps (4), (6) | Since form a locally finite open cover of (Step (4)), there exists an open subset of such that contains and intersects only finitely many of the s. Set , so is open in . therefore intersects only finitely many of the s. For any , with , we have by construction (Step (6)), so if intersects so does . Thus, the s that intersect must correspond to the finitely many s for which intersects . Since there are finitely many s for each , this gives that there are finitely many s intersecting . | ||

8 | The open cover is as desired | Steps (6), (7) | Combine the two steps to get what we wanted to prove. |