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

From Topospaces

(→Statement) |
(→Proof) |
||

Line 25: | Line 25: | ||

'''To prove''': There exists a locally finite open refinement of the <math>U_i</math>s, i.e., an open cover <math>\{ Q_k \}_{k \in K}</math> of <math>X \times Y</math> such that: | '''To prove''': There exists a locally finite open refinement of the <math>U_i</math>s, i.e., an open cover <math>\{ Q_k \}_{k \in K}</math> of <math>X \times Y</math> such that: | ||

− | * It is locally finite: | + | * It is locally finite: For any point <math>(x,y) \in X \times Y</math>, there exists an open set <math>R</math> containing <math>(x,y)</math> that intersects only finitely many of the <math>Q_k</math>s. |

* It refines <math>\{ U_i \}_{i \in I}</math>: Every <math>Q_k</math> is contained in one of the <math>U_i</math>s. | * It refines <math>\{ U_i \}_{i \in I}</math>: Every <math>Q_k</math> is contained in one of the <math>U_i</math>s. | ||

Line 46: | Line 46: | ||

|- | |- | ||

| 7 || The open cover <math>\{ Q_k \}_{k \in K}</math> of Step (6) is a locally finite open cover. In other words, for any <math>(x, y) \in X \times Y</math>, there is an open subset <math>R \ni (x,y)</math> such that <math>R</math> intersects only finite many <math>Q_k</math>s. || || || Steps (4), (6) || Since <math>\{ P_j \}_{j \in J}</math> form a locally finite open cover of <math>Y</math> (Step (4)), there exists an open subset <math>S</math> of <math>Y</math> such that <math>S</math> contains <math>y</math> and <math>S</math> intersects only finitely many of the <math>P_j</math>s. Set <math>R = X \times S</math>, so <math>R</math> is open in <math>X \times Y</math>. <math>R</math> therefore intersects only finitely many of the <math>X \times P_j</math>s. For any <math>Q_k</math>, with <math>k = (i,j)</math>, we have <math>Q_k \subseteq P_j</math> by construction (Step (6)), so if <math>Q_k</math> intersects <math>R</math> so does <math>X \times P_j</math>. Thus, the <math>Q_k</math>s that intersect <math>R</math> must correspond to the finitely many <math>j</math>s for which <math>R</math> intersects <math>X \times P_j</math>. Since there are finitely many <math>k</math>s for each <math>j</math>, this gives that there are finitely many <math>Q_k</math>s intersecting <math>R</math>. | | 7 || The open cover <math>\{ Q_k \}_{k \in K}</math> of Step (6) is a locally finite open cover. In other words, for any <math>(x, y) \in X \times Y</math>, there is an open subset <math>R \ni (x,y)</math> such that <math>R</math> intersects only finite many <math>Q_k</math>s. || || || Steps (4), (6) || Since <math>\{ P_j \}_{j \in J}</math> form a locally finite open cover of <math>Y</math> (Step (4)), there exists an open subset <math>S</math> of <math>Y</math> such that <math>S</math> contains <math>y</math> and <math>S</math> intersects only finitely many of the <math>P_j</math>s. Set <math>R = X \times S</math>, so <math>R</math> is open in <math>X \times Y</math>. <math>R</math> therefore intersects only finitely many of the <math>X \times P_j</math>s. For any <math>Q_k</math>, with <math>k = (i,j)</math>, we have <math>Q_k \subseteq P_j</math> by construction (Step (6)), so if <math>Q_k</math> intersects <math>R</math> so does <math>X \times P_j</math>. Thus, the <math>Q_k</math>s that intersect <math>R</math> must correspond to the finitely many <math>j</math>s for which <math>R</math> intersects <math>X \times P_j</math>. Since there are finitely many <math>k</math>s for each <math>j</math>, this gives that there are finitely many <math>Q_k</math>s intersecting <math>R</math>. | ||

+ | |- | ||

+ | | 8 || The open cover <math>\{ Q_k \}_{k \in K}</math> is as desired || || || Steps (6), (7) || Combine the two steps to get what we wanted to prove. | ||

|} | |} |

## Revision as of 16:27, 3 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

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:

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

## Facts used

- 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 .

## Proof

**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**:

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 with . | 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. |