Compact times metacompact implies metacompact

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:

• Compact times paracompact implies paracompact
• Compact times orthocompact implies orthocompact
• Compact times Lindelof implies Lindelof

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 $U_{i}$s. 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 $U_{i}$s.
   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.