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

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− | # [[uses::Tube lemma]]: | + | # [[uses::Tube lemma]]: Suppose <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== | ||

− | '''Given''': A compact space <math>X</math>, a paracompact space <math>Y</math>. | + | '''Given''': A compact space <math>X</math>, a paracompact space <math>Y</math>. <math>\{ U_i \}_{i \in I}</math> form an open cover of <math>X \times Y</math>. |

− | '''To prove''' | + | '''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: or 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. | ||

'''Proof''': | '''Proof''': | ||

− | + | {| class="sortable" border="1" | |

− | + | ! Step no. !! Assertion/construction !! Facts used !! Given data used !! Previous steps used !! Explanation | |

− | + | |- | |

− | + | | 1 || For any point <math>y \in Y</math>, there is a finite collection of <math>U_i</math> that cover <math>X \times \{ y \}</math> || || <math>X</math> is compact || || Since <math>X</math> is compact, the subspace <math>X \times \{ y \}</math> of <math>X \times Y</math> is also compact, so the cover by the open subsets <math>U_i</math> has a finite subcover. | |

− | + | |- | |

− | + | | 2 || For any point <math>y \in Y</math>, let <math>W_y</math> be the union of this finite collection of open subsets <math>U_i</math> as obtained in Step (1). There exists an open subset <math>V_y</math> of <math>Y</math> such that <math>y \in V_y</math> and <math>X \times V_y \subseteq W_y</math> || Fact (1) || <math>X</math> is compact || Step (1) || Follows from Fact (1), setting the <math>U</math> of Fact (1) to be <math>W_y</math>. | |

− | + | |- | |

− | + | | 3 || The open subsets <math>V_y, y \in Y</math> obtained in Step (2) form an open cover of <math>Y</math>. || || || Step (2) || By Step (2), <math>y \in V_y</math>, hence <math>X \times \{ y \} \subseteq X \times V_y</math>. Since <math>\bigcup_{y \in Y} \{ y \} = Y</math>, and <math>y \in V_y \subseteq Y</math>, we get <math>\bigcup_{y \in Y} V_y = Y</math>. | |

+ | |- | ||

+ | | 4 || There exists a locally finite open refinement <math>\{ P_j \}_{j \in J}</math> of the <math>V_y</math> in <math>Y</math> || || <math>Y</math> is paracompact || Step (3) || Step-given combination direct. | ||

+ | |- | ||

+ | | 5 || For each <math>P_j</math>, <math>X \times P_j</math> is a union of finitely many intersections <math>(X \times P_j) \cap U_i</math>, all of which are open subsets || || || Steps (1) (2), (4) || Since <math>P_j</math>s refine <math>V_y</math>s (Step (4)), there exists <math>y \in Y</math> such that <math>P_j \subseteq V_y</math>. In turn, by the definition of <math>V_y</math> (Step (2)), we have <math>X \times V_y \subseteq W_y</math>, which in turn is a union of finitely many <math>U_i</math>s (Step (1)). Thus, <math>X \times P_j</math> is contained in a union of finitely many <math>U_i</math>s, and hence, is the union of its intersection with those <math>U_i</math>s. Since <math>U_i</math> are all open, the intersections <math>(X \times P_j) \cap U_i</math> are all open. | ||

+ | |- | ||

+ | | 6 || The open subsets of the form <math>(X \times P_j) \cap U_i</math> of Step (5) form an open cover of <math>X \times Y</math> that refines the <math>U_i</math>s (note that not every combination of <math>P_j</math> and <math>U_i</math> is included -- only the finitely many <math>U_i</math>s needed as in Step (5)). We will index this open cover by indexing set <math>K subseteq I \times J</math>, and call it <math>\{ Q_k \}_{k \in K}</math>, where <math>Q_k = (X \times P_j) \cap U_i</math>. In particular, if <math>k = (i,j)</math>, then <math>Q_k \subseteq P_j</math>, and for any <math>j<math>, there are finitely many <math>k \in K</math> with <math>k = (i,j)</math>. || || || Steps (4), (5) || <math>\{ P_j \}_{j \in J}</math> cover <math>Y</math>, so <math>\{ X \times P_j \}_{j \in J}</math> cover <math>X \times Y</math>. By Step (5), <math>X \times P_j</math> is the union of the <math>(X \times P_j) \cap U_i</math>, so the latter also form an open cover of <math>X \times Y</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>. | ||

+ | |} |

## Revision as of 16:25, 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

### 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: 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: or 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 . |