# Compact times paracompact implies paracompact

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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## 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:

## Facts used

1. 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: Note that in the proof below, Step (4) has three aspects (locally finite, cover, and refinement) and different later steps use different aspects of Step (4), with the specific aspect used indicated parenthetically in the Previous steps used column.

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation Commentary
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. Begin focusing on a slice.
2 For any point , let  be the union of the 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 . Use compactness to get a tube around the slice.
3 The open subsets  obtained in Step (2) form an open cover of . Step (2) By Step (2), . Since , and , we get . Project down to the paracompact space.
4 There exists a locally finite open refinement  of the  in .  is paracompact Step (3) Step-given combination direct. Use paracompactness to get the locally finite open refinement, in the projected-down setting.
5 For each ,  is a union of finitely many intersections , all of which are open subsets. Steps (1), (2), (4) (refinement aspect) [SHOW MORE] Go back to the big space, reversing the projection step.
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 , and for any , there are finitely many  with the second coordinate of  equal to . Steps (4) (cover aspect), (5)  cover , so  cover . By Step (5),  is the union of finitely many , so the latter also form an open cover of . Wrap things up.
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) (locally finite aspect), (6) [SHOW MORE] Wrap things up.
8 The open cover  is as desired Steps (6), (7) Combine the two steps to get what we wanted to prove. --
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