Compact times paracompact implies paracompact

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 Paracompact space (?), is a topological space satisfying the property Paracompact space (?).
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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. 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:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	For any point $y\in Y$, there is a finite collection of $U_i$ that cover $X\times \{y\}$		$X$ is compact		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	For any point $y\in Y$, let $W_y$ be the union of this finite collection of open subsets $U_i$ as obtained in Step (1). There exists an open subset $V_y$ of $Y$ such that $y\in V_y$ and $X\times V_y\subseteq W_y$	Fact (1)	$X$ is compact	Step (1)	Follows from Fact (1), setting the $U$ of Fact (1) to be $W_y$.
3	The open subsets $V_y,y\in Y$ obtained in Step (2) form an open cover of $Y$.			Step (2)	By Step (2), $y\in V_y$, hence $X\times \{y\}\subseteq X\times V_y$. Since ${\bigcup }_{y\in Y}\{y\}=Y$, and $y\in V_y\subseteq Y$, we get ${\bigcup }_{y\in Y}{V}_y=Y$.
4	There exists a locally finite open refinement $\{P_j{\}}_{j\in J}$ of the $V_y$ in $Y$		$Y$ is paracompact	Step (3)	Step-given combination direct.
5	For each $P_j$, $X\times P_j$ is a union of finitely many intersections $(X\times P_j)\cap U_i$, all of which are open subsets			Steps (1) (2), (4)	Since $P_j$s refine $V_y$s (Step (4)), there exists $y\in Y$ such that $P_j\subseteq V_y$. In turn, by the definition of $V_y$ (Step (2)), we have $X\times V_y\subseteq W_y$, which in turn is a union of finitely many $U_i$s (Step (1)). Thus, $X\times P_j$ is contained in a union of finitely many $U_i$s, and hence, is the union of its intersection with those $U_i$s. Since $U_i$ are all open, the intersections $(X\times P_j)\cap U_i$ are all open.
6	The open subsets of the form $(X\times P_j)\cap U_i$ of Step (5) form an open cover of $X\times Y$ that refines the $U_i$s (note that not every combination of $P_j$ and $U_i$ is included -- only the finitely many $U_i$s needed as in Step (5)). We will index this open cover by indexing set $KsubseteqI\times J$, and call it $\{Q_k{\}}_{k\in K}$, where $Q_k=(X\times P_j)\cap U_i$. In particular, if $k=(i,j)$, then $Q_k\subseteq P_j$, and for any $j<math>,therearefinitelymany<math>k\in K$ with $k=(i,j)$.			Steps (4), (5)	$\{P_j{\}}_{j\in J}$ cover $Y$, so $\{X\times P_j{\}}_{j\in J}$ cover $X\times Y$. By Step (5), $X\times P_j$ is the union of the $(X\times P_j)\cap U_i$, so the latter also form an open cover of $X\times Y$.
7	The open cover $\{Q_k{\}}_{k\in K}$ of Step (6) is a locally finite open cover. In other words, for any $(x,y)\in X\times Y$, there is an open subset $R\ni (x,y)$ such that $R$ intersects only finite many $Q_k$s.			Steps (4), (6)	Since $\{P_j{\}}_{j\in J}$ form a locally finite open cover of $Y$ (Step (4)), there exists an open subset $S$ of $Y$ such that $S$ contains $y$ and $S$ intersects only finitely many of the $P_j$s. Set $R=X\times S$, so $R$ is open in $X\times Y$. $R$ therefore intersects only finitely many of the $X\times P_j$s. For any $Q_k$, with $k=(i,j)$, we have $Q_k\subseteq P_j$ by construction (Step (6)), so if $Q_k$ intersects $R$ so does $X\times P_j$. Thus, the $Q_k$s that intersect $R$ must correspond to the finitely many $j$s for which $R$ intersects $X\times P_j$. Since there are finitely many $k$s for each $j$, this gives that there are finitely many $Q_k$s intersecting $R$.
8	The open cover $\{Q_k{\}}_{k\in K}$ is as desired			Steps (6), (7)	Combine the two steps to get what we wanted to prove.