Lemma 6

Let $X$ be a Hausdorff space such that, given any open covering $\mathfrak{R}$ of $X$, there exists a locally finite closed covering $\mathscr{F}$ of $X$ which refines $\mathfrak{R}$. Then $X$ is paracompact.

Let $\mathfrak{R}$ be any open covering of $X$. We have to show that there is a locally finite open covering of $X$ which refines $\mathfrak{R}$. Let $\mathscr{A}$ be a locally finite covering (closed or not) of $X$ which refines $\mathfrak{R}$; for each $x \in X$, let $W_x$ be an open neighbourhood of $x$ which meets only a finite number of sets of $\mathscr{A}$. The family $\mathscr{W}$ of sets $W_x$ is an open covering of $X$. Let $\mathscr{F}$ be a locally finite closed covering of $X$ which refines $\mathscr{W}$. For each $A \in \mathscr{A}$, let $U_A$ be a set of $\mathfrak{R}$ which contains $A$, and let $C_A$ be the union of the sets $F \in \mathscr{F}$ such that $A \cap F = \varnothing$. Since $\mathscr{F}$ is locally finite, $C_A$ is closed in $X$ (Chapter I, § 1, no. 6, Proposition 4) and therefore $A' = U_A \cap (X - C_A)$ is open. Since we have $A \cap C_A = \varnothing$ and $A \subset U_A$, it follows that $A \subset A'$, and the family $\mathscr{A}'$ of sets $A'$, as $A$ runs through $\mathscr{A}$, is an open covering of $X$; moreover, since $A' \subset U_A \in \mathfrak{R}$, $\mathscr{A}'$ refines $\mathfrak{R}$. It remains to show that $\mathscr{A}'$ is locally finite. For each $x \in X$ there is a neighbourhood $T$ of $x$ which meets only a finite number of sets of $\mathscr{F}$, say $F_1, \ldots, F_n$. Since each $F_i$ is contained in a set of the form $W_{y_i}$, by definition $F_i$ meets only a finite number of sets of $\mathscr{A}$; let $A_{ij}$ ($1 \leq j \leq s_i$) be these sets. If $A$ is a set of $\mathscr{A}$ other than one of the $A_{ij}$ ($1 \leq i \leq n,\ 1 \leq j \leq s_i$), it follows from the definitions that $A'$ meets none of the $F_i$, and therefore does not meet $T \subset \bigcup_{i=1}^n F_i$. This completes the proof of Lemma 6 and of Theorem 4.