Refining open covers of a product space

Suppose $X,Y$ are topological spaces. If $\mathcal{U}$ is an open cover of $X$ and $V$ is an open cover of $Y$, let $\mathcal{U}\times \mathcal{V}$ denote the open cover of $X\times Y$ consisting of sets $U\times V$ where $U\in \mathcal{U}$ and $V\in \mathcal{V}$.

I would like to have conditions on $X$ and $Y$ so that given an open cover $\mathcal{W}$ of $X\times Y$, one can find open covers $\mathcal{U}$ of $X$ and $\mathcal{V}$ of $Y$ such that $\mathcal{U}\times \mathcal{V}$ refines $\mathcal{W}$.

It seems like $X$ and $Y$ both compact should do the trick. Am I missing something more general?

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By refine do you mean that for each $W \in \mathcal W$ and $x \in W$ there exist $U \times V \in \mathcal U \times \mathcal V$ such that $x \in U \times V \subset W$? Does the reverse hold, $U \times V \in \mathcal U \times \mathcal V$ if and only if you can find such a $W \in \mathcal W$ such that $U\times V \subset W$? – JSchlather Mar 29 '12 at 14:34
I mean refinement in the usual sense: en.wikipedia.org/wiki/Cover_(topology) – J.K.T. Mar 29 '12 at 15:04
So in this case, $\mathcal{U}\times \mathcal{V}$ refines $\mathcal{W}$ if for each $U\in \mathcal{U}$ and $V\in \mathcal{V}$, $U\times V\subseteq W$ for some $W\in \mathcal{W}$. – J.K.T. Mar 29 '12 at 15:07
Definitely...but I am asking whether any open cover can be refined with a special type of open cover that consists of rectangles. Not every open cover of $X\times Y$ consisting of rectangles is of the form $\mathcal{U}\times \mathcal{V}$ for open cover $\mathcal{U}$ of $X$ and $\mathcal{V}$ of $Y$. – J.K.T. Mar 29 '12 at 18:39

I believe this is provable with a slightly modification of the proof that the product of compact spaces is compact. Let $X,Y$ be compact spaces. It suffices to prove it for any open cover of basis elements, since we may always refine an open cover to an open cover of basis elements. Take an open cover $\mathcal W$ of $X \times Y$ be basis elements of the form $U \times V$. Cover each slice $\{x\} \times Y$ by finitely many sets in $\mathcal W$, $U_1\times V_1\dots,U_n\times V_n$. Let $W_x=\cap U_i$ then $x \in W_x$ and so $W_x$ is a non-empty open subset of $X$. Let $T_x=\{ W_x \times V_i \mid 1\leq i \leq n \}$. Now the $W_x$ form a cover of $X$ so we can find a finite subcover, $W_{x_1},\dots,W_{x_n}$. Finally for each $y \in Y$ set

$$S_y=\{V \subset Y\mid \exists i \text{ so that } W_{x_i} \times V \in T_{x_i} \text{ and } y \in V\}$$

then define $$V_y=\cap_{V \in S_y} V$$

note this is a non-empty open set and finally set $$\mathcal V=\{V_y \mid y \in Y\}$$ note this is a finite set, since there are only finitely many sets to intersect. Finally we claim that $\mathcal U \times \mathcal V$ where $\mathcal U$ is the union of the $W_{x_i}$. By construction we have ensured each product is contained inside some element of $\mathcal W$ and we can similarly see that each $x$ is contained in some $W_{x_i}$ and thereby $(x,y) \in W_{x_i} \times V_y$.

The following shows that the hypothesis of compactness is required. Consider $Z=[0,1] \times \mathbb N$ and construct the following open cover of $Z$ let

$$\mathcal U_n=\{[0,1/2+1/(n+2))\times\{n\}, (1/2,1]\times\{n\}\}$$ and

$$\mathcal U=\bigcup_{n=1}^\infty \mathcal U_n.$$

Then $\mathcal U$ is an open cover of $Z$. Suppose that $\mathcal V \times \mathcal W$ is an open refinement of $Z$. Then we must have $\mathcal W$ is the set of singletons of $\mathbb N$, since no set in $\mathcal U$ is contained in two "levels" of $\mathbb N$. Now consider $V \times \{1\} \in \mathcal V \times \mathcal W$ such that $1/2 \in V$. Then $V \subset [0,1/2+1/3)$ so let $\delta=\sup\{x \mid x \in V\}$ note that $1/2<\delta \leq 5/6$ notably we may find an $k \in \mathbb N$ such that $1/2+1/(k+3) < \delta$ and in particular we see that $V \times \{k\}$ is not contained in $[0,1/2+1/(k+2))$ or $(1/2,1]$ so it is not possible that $\mathcal V \times \mathcal W$ is a refinement of $\mathcal U$.

Edited: To remove incorrect second example.

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I believe you mean to write $\mathcal{U}_{n}$ as a set of neighborhoods in the first example. I do see how it fails for $[0,1]\times \mathbb{N}$ when $\mathbb{N}$ is discrete. And do you mean "indiscrete" in your second example? Since you are saying "compact" I'm guessing so. If this is the case, basic neighborhoods are of the form $(a,b)\times \mathbb{N}$ and the sets in $\mathcal{U}$ are not open. – J.K.T. Mar 29 '12 at 20:41
@J.T. Yeah you're right. – JSchlather Mar 29 '12 at 20:47
@J.T. Okay, I think that proves it. Unless I made some hideous mistake, besides the notation, which is definitely hideous. – JSchlather Mar 29 '12 at 21:50
Thanks for the edit, Jacob. The simple example $[0,1]\times \mathbb{N}$ confirms the unlikelihood of moving much beyond two compact factors. This is exactly what I was looking for. – J.K.T. Mar 30 '12 at 0:45