# 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 (?).
View other such computations

## 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

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 $K subseteq I \times J$, 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 $Q_k \subseteq P_j$, and for any $j[itex], there are finitely many [itex]k \in K$ 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.