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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.

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@Mark: "two such spaces" is unclear; do you mean, as your title suggests, a product of a paracompact and a compact space? Or do you mean, as "two such spaces" suggests, the product of two paracompact, or of two compact spaces? Or all of the above? – Arturo Magidin Jan 2 '11 at 23:00
@Arturo: Sorry, edited for clarity. – Mark Jan 2 '11 at 23:13
@Mark: The result as stated is false, because you are asking the paracompact space to be Hausdorff but not the compact one. Take a product of a paracompact space with an indiscrete finite space; it will be non-Hausdorff, hence not paracompact by your definition. You should either ask for Hausdorff in both, or in neither (both are definitions that are used by some). – Arturo Magidin Jan 2 '11 at 23:31
@Arturo: Why would the product of a paracompact and indiscrete space be non-Hausdorff? (Intuitively it seems that the product of a Hausdorff space and any other space should be Hausdorff) – Mark Jan 2 '11 at 23:39
@Mark: Take the underlying sets of $X$ and $Y$ to be $\{x,y\}$, give $X$ the discrete topology, $Y$ the indiscrete topology. The open sets of $X\times Y$ are $\emptyset$, $\{(x,x),(x,y)\}$, $\{(y,x), (y,y)\}$, and $X\times Y$. What are the disjoint neighborhoods of $(x,x)$ and $(x,y)$? The error in your intuition is that $(a,b)\neq(c,d)$ does not imply $a\neq c$. – Arturo Magidin Jan 2 '11 at 23:53
up vote 3 down vote accepted

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 ${\cal U}'$.

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You have the step: Now, those Wx form an open cover W of X. By hypothesis, there is a locally finite refinement W′={Wxi} Here, it seems that you are assuming a finite subcover, not a locally finite refinement. Also, for the last step, what if the set V contains infinitely many points? Then it seems it contains infinitely many slices of the form {x}*Y, and each slice intersects at least one neighborhood in your refinement. – Mark Jan 3 '11 at 16:24
Actually, I get it now, thanks! – Mark Jan 4 '11 at 1:16
@Mark. Thank you for saving my time! :-) – a.r. Jan 4 '11 at 2:14

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

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