A question about spaces having a locally finite refinement I've been reviewing questions from previous exams I've had in my courses, and I noticed a question that I completely missed. I was wondering if anyone could help me out on this one.
Let $X$ be a topological space. Prove that every $\sigma$-discrete open cover of $X$ has a locally finite refinement. 
Thanks in advance for any help!
 A: Suppose that $\mathscr{U}=\bigcup_{n\in\Bbb N}\mathscr{U}_n$ is a $\sigma$-discrete open cover of $X$, where each $\mathscr{U}_n$ is discrete. Recall (or prove!) that every discrete family is closure-preserving. That is, if $\mathscr{D}$ is a discrete family of sets, and $\mathscr{A}\subseteq\mathscr{D}$, then $\operatorname{cl}\bigcup\mathscr{A}=\bigcup\{\operatorname{cl}A:A\in\mathscr{A}\}$. (More generally, this is actually true for locally finite families of sets.) For $n\in\Bbb N$ let $U_n=\bigcup\mathscr{U}_n$, and let $V_n=\bigcup_{k\le n}U_k$.
Let $\mathscr{R}_0=\mathscr{U}_0$, and for each $n\in\Bbb N$ let $\mathscr{R}_{n+1}=\{U\setminus V_n:U\in\mathscr{U}_{n+1}\}$. Let $\mathscr{R}=\bigcup_{n\in\Bbb N}\mathscr{R}_n$. Clearly $\mathscr{R}$ refines $\mathscr{U}$. An easy induction verifies that $\bigcup\bigcup_{k\le n}\mathscr{R}_k=V_n$ for each $n\in\Bbb N$, so $\mathscr{R}$ covers $X$. To see that $\mathscr{R}$ is locally finite, fix $x\in X$. Let $n$ be minimal such that $x\in U_n$, and let $U$ be the unique member of $\mathscr{U}_n$ containing $x$. Suppose that $R\in\mathscr{R}$, and $U\cap R\ne\varnothing$; $U\subseteq V_n$, and $V_n\cap\bigcup\mathscr{R}_k=\varnothing$ for $k>n$, so $R\in\mathscr{R}_k$ for some $k\le n$. For each $k\le n$, $x$ has an open nbhd $B_k$ that meets at most one member of $\mathscr{U}_k$ and hence also at most one member of $\mathscr{R}_k$. Let $B=U\cap\bigcap_{k\le n}B_k$; then $B$ is an open nbhd of $x$ that meets at most finitely many members of $\bigcup_{k\le n}\mathscr{R}_k$ and no members of $\bigcup_{k>n}\mathscr{R}_k$, so $B$ meets only finitely many members of $\mathscr{R}$. Since $x$ was arbitrary, $\mathscr{R}$ is locally finite.
Of course $\mathscr{R}$ is almost certainly not an open refinement of $\mathscr{U}$. To get that, you have to impose some conditions on $X$, and you have to work harder. To see that some extra conditions are definitely necessary, consider the following example.
Let $\tau$ be the following $T_1$ topology on $\Bbb R$. First, let $D=\Bbb R\setminus\Bbb Z^+$; points of $D$ are isolated. Each $n\in\Bbb Z^+$ has basic open nbhds of the form $\{n\}\cup(D\setminus F)$, where $F$ is any finite subset of $D$. Then $$\mathscr{U}=\{D\}\cup\Big\{\{n\}\cup D:n\in\Bbb Z^+\Big\}$$ is a countable and hence $\sigma$-discrete open cover of $X$ with no locally finite open refinement: any open refinement will have to include a nbhd of $n$ for each $n\in\Bbb Z^+$, and these nbhds will necessarily have non-empty intersection, since their intersection can exclude only countably many members of the uncountable set $D$.
