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: 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.

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation	Commentary
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.	Begin focusing on a slice.
2	For any point $y\in Y$, let $W_{y}$ be the union of the 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 $W_{y}$.	Use compactness to get a tube around the slice.
3	The open subsets $V_{y},y\in Y$ obtained in Step (2) form an open cover of $Y$.			Step (2)	By Step (2), $y\in V_{y}$. Since $\bigcup _{y\in Y}\{y\}=Y$, and $y\in V_{y}\subseteq Y$, we get $\bigcup _{y\in Y}V_{y}=Y$.	Project down to the paracompact space.
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.	Use paracompactness to get the locally finite open refinement, in the projected-down setting.
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) (refinement aspect)	[SHOW MORE] 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.	Go back to the big space, reversing the projection step.
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 $K\subseteq I\times J$, 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 $Q_{k}\subseteq X\times P_{j}$ and $Q_{k}\subseteq U_{i}$, and for any $j$, there are finitely many $k\in K$ with the second coordinate of $k$ equal to $j$.			Steps (4) (cover aspect), (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 finitely many $(X\times P_{j})\cap U_{i}$, so the latter also form an open cover of $X\times Y$.	Wrap things up.
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) (locally finite aspect), (6)	[SHOW MORE] 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 $Q_{k}\subseteq X\times P_{j}$ 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 $R$.	Wrap things up.
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.	--

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