Htykuut

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.

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 ${cal W}' = left{ W_{x_i}right}$, $iin I$ for some index set $I$.

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

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

with $iin I$ 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 $Vtimes Y$ is a neighborhood of $(x,y)$ that intersects only finitely many of the sets of ${cal U}'$.

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 $Xtimes Y$.

(Tube lemma part) First fix $xin X$, and for each $y in Y$, find $Ainmathcal{A}$ and basis element $Utimes V$ such that $(x,y)in Utimes Vsubseteq A$. As $y$ ranges in $Y$, these various $Utimes V$ cover ${x}times Y$, which is compact. Thus there exists finitely many $U_1times V_1 subseteq A_1,dots,U_ntimes V_nsubseteq A_n$ that cover ${x}times Y$. Let $U_x = U_1cap dots cap U_n$. For later use, let $mathcal{A}_x={A_1,...,A_n}$.

Now, ${U_x}_{xin X}$ 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 $B_i,iin I$ are the elements of $mathcal{B}$. Using the refinement property, for each $iin I$, pick $x_iin X$ such that $B_isubseteq U_{x_i}$.

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

$$mathcal{C_{x_i}}:={Acap (B_itimes Y)}_{Ain mathcal{A}_{x_i}},quad mathcal{C}:=bigcup_{iin I}mathcal{C}_{x_i}$$

To prove that this is a cover, consider any $(x,y)in Xtimes Y$. 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 $(x,y)in Xtimes Y$. First there exists an open neighbourhood $Usubseteq X$ of $x$ that intersects only finitely many elements of $mathcal{B}$, say $B_1,...,B_m$. Then $Utimes Y$ 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.

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