The product of a paracompact space and a compact space is paracompact. (Why?)

Question

A paracompact space is a space in which every open cover has a locally finite refinement.

A compact space is a space in which every open cover has a finite subcover.

Why must the product of a compact and a paracompact space be paracompact?

I really have very little intuition about how to go about this question, so any hints or a proof would be greatly appreciated.

Answer 1

The key idea is indeed to use the "tube lemma", as d.t. did in his solution, but there is a small gap in his solution.

(This proof does not assume prior knowledge of tube lemma)

Let $X$ be paracompact and $Y$ be compact. Let $\mathcal{A}$ be an open cover of $X\times Y$.

(Tube lemma part) First fix $x\in X$, and for each $y \in Y$, find $A\in\mathcal{A}$ and basis element $U\times V$ such that $(x,y)\in U\times V\subseteq A$. As $y$ ranges in $Y$, these various $U\times V$ cover $\{x\}\times Y$, which is compact. Thus there exists finitely many $U_1\times V_1 \subseteq A_1,\dots,U_n\times V_n\subseteq A_n$ that cover $\{x\}\times Y$. Let $U_x = U_1\cap \dots \cap U_n$. For later use, let $\mathcal{A}_x=\{A_1,...,A_n\}$.

Now, $\{U_x\}_{x\in X}$ is an open cover of $X$. Using paracompactness, there is an locally finite open refinement $\mathcal{B}$ that covers $X$. For purpose of notation, suppose $B_i,i\in I$ are the elements of $\mathcal{B}$. Using the refinement property, for each $i\in I$, pick $x_i\in X$ such that $B_i\subseteq U_{x_i}$.

Consider the open refinement $\mathcal{C}$ of $\mathcal{A}$ given by

$$\mathcal{C_{x_i}}:=\{A\cap (B_i\times Y)\}_{A\in \mathcal{A}_{x_i}},\quad \mathcal{C}:=\bigcup_{i\in I}\mathcal{C}_{x_i}$$

To prove that this is a cover, consider any $(x,y)\in X\times Y$. First $x$ is in some $B_i$. Since $\mathcal{C}_{x_i}$ covers $B_i \times Y$, $(x,y)$ is covered by $\mathcal{C}$.

To prove that it is locally finite, consider any $(x,y)\in X\times Y$. First there exists an open neighbourhood $U\subseteq X$ of $x$ that intersects only finitely many elements of $\mathcal{B}$, say $B_1,...,B_m$. Then $U\times Y$ is the desired neighbourhood of $(x,y)$ that only intersects finitely many elements of $\mathcal{C}$ as it can only intersect elements from $\mathcal{C}_{x_1},...,\mathcal{C}_{x_m}$, each of which is a finite collection.

Answer 2

I think Trevor is right. Here are the details (this is an adaptation of the classical proof that the product of two compact spaces is a compact space that you can find in Munkres, for instance).

Let $X$ be a paracompact space, $Y$ a compact one and ${\cal U}$ an open cover of $X \times Y$.

For any $x \in X$, the slice $\left\{ x \right\} \times Y$ is a compact space and ${\cal U}$ an open cover of it (as a subspace). So it admits a finite subcover $U_{x,1}, \dots , U_{x,n_x} \in {\cal U}$. Let us call $N_x = U_{x,1} \cup \dots \cup U_{x,n_x}$ their union.

So $N_x$ is an open set that contains the slice $\left\{ x \right\} \times Y$. Because of the tube lemma, there exists an open set $W_x \subset X$ such that

$$ \left\{ x \right\} \times Y \quad \subset \quad W_x \times Y \quad \subset \quad N_x \ . $$

Now, those $W_x$ form an open cover ${\cal W}$ of $X$. By hypothesis, there is a locally finite refinement ${\cal W}' = \left\{ W_{x_i}\right\}$, $i\in I$ for some index set $I$.

Consider the following subcover of ${\cal U}$:

$$ {\cal U}' = \left\{ U_{x_i,j}\right\} \ , $$

with $i\in I$ and $j = 1, \dots , n_{x_i}$.

Let us show that ${\cal U}'$ is a locally finite subcover of ${\cal U}$: take any point $(x,y) \in X \times Y$. By hypothesis, there is a neighborhood $V \subset X$ of $x$ such that $V$ interesects only finitely many of the sets of ${\cal W}'$. Then $V\times Y$ is a neighborhood of $(x,y)$ that intersects only finitely many of the sets of ${\cal U}'$.

Answer 3

You could prove that the product of a paracompact space and a compact space is paracompact by using the tube lemma.