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 (?).
View other such computations
Other results using the same proof technique:
- Compact times metacompact implies metacompact
- Compact times orthocompact implies orthocompact
- Compact times Lindelof implies Lindelof
- 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.
|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 , there are finitely many with the second coordinate of equal to .||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.|