Compact times paracompact implies paracompact

From Topospaces
Revision as of 03:13, 17 July 2009 by Vipul (talk | contribs)

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

Statement

Verbal statement

The product of a compact space with a paracompact space (given the product topology), is paracompact

Statement with symbols

Let be a compact space and a paracompact space. Then is paracompact.

Related facts

Other results using the same proof technique:

Facts used

  1. 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 .

Proof

Given: A compact space , a paracompact space .

To prove If form an open cover of , there exists a locally finite open refinement of the .

Proof:

  1. 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.
  2. Let be the union of this finite collection of open subsets . By fact (1), there exists an open subset of such that .
  3. The form an open cover of .
  4. There exists a locally finite open refinement, say of the in : This follows from the fact that is paracompact.
  5. We can construct a locally finite open refinement of from these:
    1. 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 .
    2. Since the together cover , the are an open cover of that refines the s.
    3. Finally, we argue that is a locally finite open cover: Suppose . Since is a locally finite open cover of , there exists an open subset of containing such that intersects only finitely many members of . Thus, the neighborhood intersects only finitely many s, which in turn give rise to finitely many s each. Thus, intersects only finitely many members of the open cover.