# 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:

## 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.