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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?2011-01-02
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    @Arturo: Sorry, edited for clarity.2011-01-02
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    @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).2011-01-02
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    @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)2011-01-02
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    Haha, they named a topology after me: the indiscreet topology.2011-01-02
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    @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 $a\neq c$.2011-01-02
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    @Arturo: That is a good counter-example to the question as stated. It looks like X is paracompact and Y is compact, and yet their product is non-Hausdorff. I will remove the requirement that paracompactness -> Hausdorff.2011-01-03
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    @Mark: Yes, both $X$ and $Y$ are compact, hence $X$ is paracompact. A product of two spaces (with the product topology) is Hausdorff if and only if both spaces are Hausdorff, so you can either add the condition in *both*, or drop it in both. Many authors require Hausdorff-ness for compactness, and many authors do not require it for paracompactness, so it's six of one, half a dozen of the other.2011-01-03

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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.2011-01-03
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    Actually, I get it now, thanks!2011-01-04
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    @Mark. Thank you for saving my time! :-)2011-01-04
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    Are you sure the last sentence is correct? I fail to see how this holds true (might completely be on me, I'd just like to understand, then): From what I understand all we can conclude is that $V \times Y$ is *contained* in finitely many of the sets of $\mathcal{U}'$; however, I don't get how we can assume that other elements of $\mathcal{U}'$ have empty intersection with $V \times Y$. I'm thinking of $W_x \subsetneq N_x\,,\;x \in X$...2017-06-01
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    Just found a proof at https://topospaces.subwiki.org/wiki/Compact_times_paracompact_implies_paracompact ; there an important last step is included, which I would say is missing here. However, they are a bit sloppy in point 5, as they miss to state explicitly that they are only looking at finitely many intersections for each $P$.2017-06-01
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    I agree with polynomial_donut; I believe there is an issue with this proof. After the sentence "By hypothesis, there is a locally finite refinement $W′=\{W_{x_i}\}, i\in I$ for some index set $I$", I believe you cannot just take the subcover $\mathcal{U}'$ that is in your proof. Rather, you should consider the refinement $\mathcal{U}' :=\{U_{x_i,j} \cap (W_{x_i}\times Y)\}$. Without the additional intersection, the last sentence of your proof does not follow from the 2nd-to-last sentence.2017-09-30
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    The fix I proposed also misses a step, so I decided I should write up a solution instead2017-09-30
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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.

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