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 \(\ts{B_\alpha}_{\alpha \in J}\) be a locally finite indexed family of subsets of the paracompact Hausdorff space \(X\). Then there is a locally finite indexed family \(\ts{U_\alpha}_{\alpha \in J}\) of open sets in \(X\) such that \(B_\alpha \subset U_\alpha\) for each \(\alpha\).
Solution
For each \(x\in X\), local finiteness of \(\{B_\alpha\}\) gives an open neighborhood \(N_x\) meeting only finitely many \(B_\alpha\)’s. The family \(\{N_x:x\in X\}\) is an open cover of the paracompact space \(X\), so choose a locally finite open refinement \(\mathcal V\) covering \(X\), with each \(V\in\mathcal V\) contained in some \(N_x\). In particular, each \(V\in\mathcal V\) meets only finitely many \(B_\alpha\)’s.
For each \(\alpha\), define \[ U_\alpha=\bigcup\{V\in\mathcal V:V\cap B_\alpha\ne\varnothing\}. \] Then \(U_\alpha\) is open. It contains \(B_\alpha\): if \(b\in B_\alpha\), some \(V\in\mathcal V\) contains \(b\), and that \(V\) occurs in the displayed union.
It remains to prove local finiteness of \(\{U_\alpha\}\). Fix \(x\in X\). Since \(\mathcal V\) is locally finite, choose an open neighborhood \(W\) of \(x\) meeting only \(V_1,\ldots,V_k\in\mathcal V\). If \(W\cap U_\alpha\ne\varnothing\), then some \(V_i\) meeting \(W\) also meets \(B_\alpha\). Each \(V_i\) meets only finitely many \(B_\alpha\)’s, so only finitely many indices \(\alpha\) can occur. Thus \(W\) meets only finitely many \(U_\alpha\)’s.
Hence \(\{U_\alpha\}_{\alpha\in J}\) is a locally finite indexed family of open sets with \(B_\alpha\subset U_\alpha\) for every \(\alpha\).