Lemma 3

*Let $\mathscr{R} = (U_\alpha)_{\alpha \in A}$ be an open covering of a metrizable space X. Then there is a sequence $(\mathscr{G}_n)$ of locally finite families of open subsets of X, such that $\mathscr{S} = \bigcup_n \mathscr{S}_n$ is an open covering of X which refines $\mathscr{R}$.

Let $d$ be a metric on X compatible with its topology. For each $\alpha \in A$ and each integer $n$, let $F_{n\alpha}$ denote the set of all $x \in U_\alpha$ such that $d(x, X - U_\alpha) \geq 2^{-n}$. Since $X - U_\alpha$ is closed, we have $U_\alpha = \bigcup_n F_{n\alpha}$.

Well-order the set $A$; for each $\alpha \in A$ and each integer $n$, let $G_{n\alpha}$ be the set of all $x \in F_{n\alpha}$ such that $x \notin F_{n+1,\beta}$ for all $\beta < \alpha$, and let $V_{n\alpha}$ be the set of all $y \in X$ such that $d(y, G_{n\alpha}) > 2^{-n-3}$. $V_{n\alpha}$ is clearly an open set; on the other hand, $V_{n\alpha} \subset U_\alpha$, because for each $y \in V_{n\alpha}$ there exists $x \in G_{n\alpha}$ such that $d(x, y) \leq 2^{-n-1}$, and since $x \in F_{n\alpha}$, we have

so that $y \in U_\alpha$. Furthermore, for each $x \in X$ let $\alpha$ be the smallest index in $A$ such that $x \in U_\alpha$; then there is an integer $n$ such that $x \in F_{n\alpha}$, and the definition of $\alpha$ implies that $x \in G_{n\alpha}$, so that $x \in V_{n\alpha}$. This shows that if we put $\mathfrak{S}_n = (V_{n\alpha})_{\alpha \in A}$, then $\mathfrak{S} = \bigcup_n \mathfrak{S}_n$ is an open covering of $X$ which refines $\mathscr{R}$; thus it remains to be shown that each of the families $\mathfrak{S}_n$ is locally finite. To this end we shall first show that $d(G_{n\alpha}, G_{n\beta}) \geq 2^{-n-1}$ if $\alpha \neq \beta$. Suppose that $\beta < \alpha$; then if $x \in G_{n\alpha}$ and $y \in F_{n\beta}$ we have $x \notin F_{n+1,\beta}$ by definition, hence $d(x, X - U_\beta) < 2^{-n-1}$ and $d(y, X - U_\beta) \geq 2^{-n}$, and therefore $d(x, y) \geq 2^{-n-1}$; since $G_{n\beta} \subset F_{n\beta}$, the assertion follows. From this it follows immediately, using the definition of $V_{n\alpha}$ and $V_{n\beta}$, that $d(V_{n\alpha}, V_{n\beta}) \geq 2^{-n-2}$. From this last inequality we deduce that, for each $z \in X$, the open ball with centre $z$ and radius $2^{-n-3}$ meets at most one set of the family $\mathfrak{S}_n$; hence $\mathfrak{S}_n$ is a locally finite family, and the proof is complete.