Compact times metacompact implies metacompact

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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 Metacompact space (?), is a topological space satisfying the property Metacompact space (?).
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Statement

Verbal statement

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

Statement with symbols

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

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 X \times V \subseteq U.

Proof

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

To prove If U_i form an open cover of X \times Y, there exists a point-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 point-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 point-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 point-finite open cover: Suppose (x,y) \in X \times Y. Since \mathcal{P} is a point-finite open cover of Y. Then, there exist only finitely many P \in \mathcal{P} such that y \in P. For each of these, X \times P corresponds to finitely many intersections (X \times P) \cap U_i, so the total number of open subsets containing (x,y) is finite.