In the paper by Ernest Michael cited in the title of this question, the following Lemma is crucial, but a step in the proof confuses me.
Lemma 1. Let $X$ be a regular space. Then the following are equivalent:
- Every open covering of $X$ has an open locally finite refinement, i.e. $X$ is paracompact.
- Every open covering of $X$ has a locally finite refinement.
- Every open covering of $X$ has a closed locally finite refinement.
Here is the proof given for (3) $\implies$ (1), copied mostly verbatim, with one annotation of the form [?!] for reference below.
Proof. Let $\mathcal{R}$ be an open covering of $X$. Let $\mathcal{A}$ be a locally finite refinement of $\mathcal{R}$, and let $\mathcal{B}$ be a covering of $X$ by open sets, each of which intersects only finitely many elements of $\mathcal{A}$. Now let $\mathcal{W}$ be a closed, locally finite refinement of $\mathcal{B}$. For each $A \in \mathcal{A}$, let $A' = X \setminus \bigcup \{ W \in \mathcal{W} \, | \, A \cap W = \emptyset \}$; then $A'$ is an open set [?!] containing $A$, and if $W \in \mathcal{W}$, then $W$ intersects $A'$ if and only if $W$ intersects $A$. For each $A \in \mathcal{A}$, pick an $R_A \in \mathcal{R}$ such that $A \subset R_A$. Let $\mathcal{U} = \{ A' \cap R_A \, | \, A \in \mathcal{A} \}$. Then $\mathcal{U}$ is an open refinement of $\mathcal{R}$, and since each element of the locally finite covering $\mathcal{W}$ intersects only finitely many elements of $\mathcal{U}$, $\mathcal{U}$ is locally finite. $\square$
Discussion. Let $\mathcal{W}_A := \{ W \in \mathcal{W} \, | \, A \cap W = \emptyset \}$. This is a family of closed sets. As [?!] indicates, I don't see why $A' = X \setminus \bigcup \mathcal{W}_A$ is open, or equivalently why $\bigcup \mathcal{W}_A$ is closed. My confusion is that, in general, $\mathcal{W}_A$ is infinite. Indeed, note that the family of open sets $\mathcal{B}_A := \{ B \in \mathcal{B} \, | \, A \cap B = \emptyset \}$ is cofinite in $\mathcal{B}$ by assumption. If we choose a map $b: \mathcal{W} \to \mathcal{B}$ such that $W \subseteq b(W)$ for each $W \in \mathcal{W}$, then $b^{-1}(\mathcal{B}_A) \subseteq \mathcal{W}_A$, so $\mathcal{W}_A$ is infinite for every $A$ whenever $\mathcal{B}$ is infinite. But then I don't see why the union $\bigcup \mathcal{W}_A$ should be closed, since in general an infinite union of closed sets is not closed.
My first thought is to define $\mathcal{B}'_A := \{ b(W) \, | \, W \in \mathcal{W}, \, W \cap A = \emptyset \}$ (keeping the chosen map $b$ from the previous paragraph), then instead of $A'$ use $A'' := X \setminus \bigcup \mathcal{B}'_A$. Since each $b(W)$ is open, $A''$ is certainly closed, but now it's not at all clear that $A''$ contains $A$, since $b(W)$ can intersect $A$ even when $W$ does not. So this first thought is not quite right.
Stepping outside the (compact) text into the para(compact)text, I don't actually think that there is an error, because if there was one, someone would have caught it by now. (In particular, Jean Dieudonné wrote the review that is now in MathSciNet, and I trust his understanding rather more than my own.) Any indication of where I failed to understand the proof would be appreciated.
Best Answer
You are right, the family $\mathcal{W}_A$ may be infinite. But it is a locally finite family of closed sets and thus the union of its members is closed. See here or Closure of union of locally finite collection of subsets equal to union of closures.