Compact times paracompact implies paracompact

Statement
Let $$X$$ be a compact space and $$Y$$ a paracompact space. Then $$X \times Y$$, the Cartesian product endowed with the product topology, is paracompact.

Related facts
Other results using the same proof technique:


 * Compact times metacompact implies metacompact
 * Compact times orthocompact implies orthocompact
 * Compact times Lindelof implies Lindelof

Facts used

 * 1) uses::Tube lemma: Suppose $$X$$ 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 $$y \in V$$ and $$X \times V \subseteq U$$.

Proof
Given: A compact space $$X$$, a paracompact space $$Y$$. $$\{ U_i \}_{i \in I}$$ form an open cover of $$X \times Y$$.

To prove: There exists a locally finite open refinement of the $$U_i$$s, i.e., an open cover $$\{ Q_k \}_{k \in K}$$ of $$X \times Y$$ such that:
 * It is locally finite: For any point $$(x,y) \in X \times Y$$, there exists an open set $$R$$ containing $$(x,y)$$ that intersects only finitely many of the $$Q_k$$s.
 * It refines $$\{ U_i \}_{i \in I}$$: Every $$Q_k$$ is contained in one of the $$U_i$$s.

Proof: Note that in the proof below, Step (4) has three aspects (locally finite, cover, and refinement) and different later steps use different aspects of Step (4), with the specific aspect used indicated parenthetically in the Previous steps used column.