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 $⋃_{y \in Y} {y} = Y$ , and $y \in V_{y} \subseteq Y$ , we get $⋃_{y \in Y} V_{y} = Y$ .
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 $K \subseteq I \times J$ , 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), (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 $X \times Y$ .
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 $R ∋ (x, y)$ 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 $Q_{k} \subseteq X \times P_{j}$ 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 $R$ .
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.