Compact times metacompact implies metacompact
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 Metacompact space (?), is a topological space satisfying the property Metacompact space (?).
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Statement with symbols
Other results using the same proof technique:
- Compact times paracompact implies paracompact
- Compact times orthocompact implies orthocompact
- Compact times Lindelof implies Lindelof
- 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 .
Given: A compact space , a metaacompact space .
To prove If form an open cover of , there exists a point-finite open refinement of the .
- 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 point-finite open refinement, say of the in : This follows from the fact that is paracompact.
- We can construct a point-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 point-finite open cover: Suppose . Since is a point-finite open cover of . Then, there exist only finitely many such that . For each of these, corresponds to finitely many intersections , so the total number of open subsets containing is finite.