Munkres section 41, expansion lemma and paracompact spaces

Question

[Ex5] Let $X$ be paracompact. We proved a "shrinking lemma" for arbitrary indexed open coverings of $X$. Here is an "expansion lemma" for arbitrary locally finite indexed families in $X$.

Lemma. Let $\{ B_ \alpha \} _ { \alpha \in J}$ be a locally finite indexed family of subsets of the paracompact Hausdorff space $X$. Then there is a locally finite indexed family $\{ U_ \alpha \} _ { \alpha \in J}$ of open sets in $X$ such that $ B_ \alpha \subset U_ \alpha$ for each $\alpha$.

I read Brian M. Scott's answer in Expansion lemma for paracompact hausdorff space. and Showing a uniformity is complete. But I couldn't understand it because I haven't studied about barycentric refinement, star refinement, uniform spaces, completeness, and filters(also they haven't appeared in Munkres text yet). Are there any other ways to prove this?

Answer 1

Here is an answer which uses only the basic definition of paracompactness and does not require a Hausdorff assumption. There are some gymnastic involved but it's not too messy.

Let $\mathcal{A}$ be a locally-finite family of closed subsets of a paracompact space $X$. Since the closures of the members of a locally-finite family themselves form a locally-finite family, there is no loss of generality in posing the additional assumption.

Notation: Write $Fin(\mathcal{A})$ for the set of finite nonempty subsets $\alpha=\{A_1,\dots,A_n\}\subset\mathcal{A}$. For a subset $U\subseteq X$ write $\mathcal{A}_U=\{A\in\mathcal{A}\mid A\cap U\neq\emptyset\}$. For $x\in X$ we write $\mathcal{A}_x=\mathcal{A}_{\{x\}}$. Since a locally-finite family is point finite, we have $\mathcal{A}_x\in Fin(\mathcal{A})$ for each $x\in X$.

Claim: There is a locally-finite family $\mathcal{U}=\{U_A\}_{A\in\mathcal{A}}$ of open subsets of $X$ such that $A\subset U_A$ for each $A\in\mathcal{A}$.

Proof: For $\alpha\in Fin(\mathcal{A})$ write $V(\alpha)=X\setminus\bigcup(\mathcal{A}\setminus\alpha)$. By the local-finiteness of $\mathcal{A}$, each $V(\alpha)$ is open. Moreover, if $x\in X$, then $x\in V(\mathcal{A}_x)$, and thus the family $\{V(\alpha)\}_{\alpha\in Fin(\mathcal{A})}$ is an open covering of $X$. Let $\mathcal{V}$ be a locally-finite open refinement of it.

Now for $A\in\mathcal{A}$ put $U_A=St(A;\mathcal{V})=\bigcup\{V\in\mathcal{V}\mid V\cap A\neq\emptyset\}$. Then $A\subseteq U_A$ and $U_A$ is open since the sets in $\mathcal{V}$ are. We claim that $\mathcal{U}=\{U_A\}_{A\in\mathcal{A}}$ is locally-finite. In showing this we will complete the proof.

So fix a point $x\in X$. Let $W\subseteq X$ be an open neighbouhood of $x$ which meets only finitely many sets in $\mathcal{V}$, say $V_1,\dots,V_n$. For each $i=1,\dots,n$ choose $\alpha_i\in Fin(\mathcal{A})$ such that $V_i\subseteq V(\alpha_i)$ and write $\alpha=\{A\in\mathcal{A}\mid W\cap U_A\neq\emptyset\}$. Then $\alpha\subset\bigcup_{i\leq n}\alpha_i$. In particular $\alpha\in Fin(\mathcal{A})$, which shows that $W$ meets only finitely many sets in $\mathcal{U}$. $\square$