# Difference between revisions of "Compact times paracompact implies paracompact"

From Topospaces

(→Facts used) |
|||

Line 24: | Line 24: | ||

==Facts used== | ==Facts used== | ||

− | # [[uses::Tube lemma]]: If <math>X</math> is a compact space and <math>Y</math> is a topological space. Then, given any open subset <math>U</math> of <math>X \times Y</math> containing <math>X \times \{ y \}</math> for some <math>y \in Y</math>, there exists an open subset <math>V</math> of <math>Y</math> such that <math>X \times V \subseteq U</math>. | + | # [[uses::Tube lemma]]: If <math>X</math> is a compact space and <math>Y</math> is a topological space. Then, given any open subset <math>U</math> of <math>X \times Y</math> containing <math>X \times \{ y \}</math> for some <math>y \in Y</math>, there exists an open subset <math>V</math> of <math>Y</math> such that <math>y \in V</math> and <math>X \times V \subseteq U</math>. |

==Proof== | ==Proof== |

## Revision as of 14:52, 2 June 2017

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

## Contents

## 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:

- Compact times metacompact implies metacompact
- Compact times orthocompact implies orthocompact
- Compact times Lindelof implies Lindelof

## Facts used

- 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 and .

## 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**:

- 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 locally finite open refinement, say of the in : This follows from the fact that is paracompact.
- We can construct a locally 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
*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.