Suppose $X$ is a regular, Hausdorff space and that every open cover of $X$ has a locally finite refinement (not necessarily open or closed).
Let $\mathcal{U}$ be an open cover of $X$. I want to find a locally
finite closed refinement of $\mathcal{U}$, $\{F_{U} : U \in
\mathcal{U} \}$, such that for each $U \in \mathcal{U}$ we'd have $F_U
\subset U$.
The way I started is this:
For each $x \in X$ there's $U_x \in \mathcal{U}$, so by regularity we can take $x \in V_x \subset \overline{V_x} \subset U_x$ where $V_x$ is open.
Take a locally finite refinement $\mathcal{V}$ of $\{V_x:x \in X\}$.
Now $\mathcal{V_c} = \{\overline{V} : V \in \mathcal{V}\}$ is still a locally finite refinement of $\mathcal{U}$.
For each $U \in \mathcal{U}$ define $V_U = \bigcup\{ \overline{V} : V \in \mathcal{V}, V \subset U \}$. This collection clearly covers $X$ and is closed, but I don't think it is necessarily locally finite.
Can this be made to work? If not, suggest a way to restart?
Best Answer
You're pretty close, idea-wise:
I'll use indices, so let $\mathcal{U}=\{U_i: i \in I\}$ ($I$ some index set; but you can also index each set by itself if you prefer) be a cover of $X$ with some locally finite closed refinement (also a cover) $\mathcal{W}$ (no indices yet).
For every $W \in \mathcal{W}$ we can find $f(W) \in I$ such that $W \subseteq U_{f(W)}$, by the definition of refinement. This defines (using AC): $f: \mathcal{W} \to I$.
Now define $W_i = \bigcup \{W: f(W)=i\}$, which is closed as a locally finite union of closed sets. As all $W$ with $f(W)=i$ are a subset of $U_i$ by definition of $f$, $W_i \subseteq U_i$ is obvious too.
Now, $\{W_i: i \in I\}$ is also locally finite: let $x \in x$ be arbitrary and let $N_x$ be a neighbourhood of it such that $\operatorname{i}(N_x,\mathcal{W}) = \{W \in \mathcal{W}: N_x \cap W \neq \emptyset\}$ is finite.
If $N_x \cap W_i$ is non-empty, this means exactly that $N_x \cap W$ is non-empty for some $W \in \mathcal{W}$ with $f(W)=i$. It follows that
$$|\{W \in \mathcal{W}: N_x \cap W \neq \emptyset\}|= |\{W_i: W_i \cap N_x\}|$$
and so $N_x$ also witnesses that $\{W_i: i \in I\}$ is finite.
So if $\{U_i: i \in I\}$ has a locally finite closed refining cover, we can find one with the same indexing and such that $W_i \subseteq U_i$ for all $i$.