Compact times paracompact implies paracompact

From Topospaces
Jump to: navigation, search
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 (?).
View other such computations


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 .


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. --
This proof uses a tabular format for presentation. Learn more about tabular proof formats|View all pages on facts with proofs in tabular format