Compact times metacompact implies metacompact

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:


 * Compact times paracompact implies paracompact
 * Compact times orthocompact implies orthocompact
 * Compact times Lindelof implies Lindelof

Facts used

 * 1) uses::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:
 * 6) 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 $$U_i$$s. Thus, $$X \times P$$ is the union of the intersections $$(X \times P) \cap U_i$$.
 * 7) 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 $$U_i$$s.
 * 8) 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.