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:

Facts used

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 $j$ , there are finitely many $k\in K$ with the second coordinate of $k$ equal to $j$ .			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.