Compact times paracompact implies paracompact

From Topospaces

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×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×Y containing X×{y} for some yY, there exists an open subset V of Y such that yV and X×VU.

Proof

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

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

Proof:

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