Difference between revisions of "Compact times paracompact implies paracompact"

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==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>.
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# [[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 (?).
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Statement

Verbal statement

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

Statement with symbols

Let X be a compact space and Y a paracompact space. Then X \times Y is paracompact.

Related facts

Other results using the same proof technique:

Facts used

  1. Tube lemma: If X is a compact space and Y is a topological space. Then, given any open subset U of X \times Y containing X \times \{ y \} for some y \in Y, there exists an open subset V of Y such that y \in V and X \times V \subseteq U.

Proof

Given: A compact space X, a paracompact space Y.

To prove If U_i form an open cover of X \times Y, there exists a locally finite open refinement of the U_i.

Proof:

  1. For any point y \in Y, there is a finite collection of U_i that cover X \times \{ y \}: Since X is compact, the subspace X \times \{ y \} of X \times Y is also compact, so the cover by the open subsets U_i has a finite subcover.
  2. Let W_y be the union of this finite collection of open subsets U_i. By fact (1), there exists an open subset V_y of Y such that X \times V_y \subseteq W_y.
  3. The V_y form an open cover of Y.
  4. There exists a locally finite open refinement, say \mathcal{P} of the V_y in Y: This follows from the fact that Y is paracompact.
  5. We can construct a locally finite open refinement of U_i from these:
    1. For each member P \in \mathcal{P}, there exists V_y such that P \subseteq V_y. Thus, X \times P \subseteq X \times V_y \subseteq W_y. W_y, in turn, is a union of a finite collection of U_is. Thus, X \times P is the union of the intersections (X \times P) \cap U_i.
    2. Since the X \times P together cover X \times Y, the (X \times P) \cap U_i are an open cover of X \times Y that refines the U_is.
    3. Finally, we argue that (X \times P) \cap U_i is a locally finite open cover: Suppose (x,y) \in X \times Y. Since \mathcal{P} is a locally finite open cover of Y, there exists an open subset Q of Y containing y such that Q intersects only finitely many members of \mathcal{P}. Thus, the neighborhood X \times Q intersects only finitely many X \times Ps, which in turn give rise to finitely many (X \times P) \cap U_is each. Thus, X \times Q intersects only finitely many members of the open cover.