Lemma 4

Let $(\mathfrak{S}_n)$ be a sequence of locally finite families of open sets in a topological space $X$, such that

is a covering of $X$. Then there exists a locally finite (but not necessarily open) covering $\mathscr{B}$ of $X$ which refines $\mathfrak{S}$.

Let $E_n$ be the open set in $X$ which is the union of all the sets of $\mathfrak{S}_n$; let $U_n$ denote $\bigcup_{k=1}^n E_k$ and let $A_n$ denote $U_n - U_{n-1}$ (with $U_0 = \varnothing$). Consider the set $\mathscr{B}$ of subsets $V \cap A_n$, where $V \in \mathfrak{S}_n$ and $n$ is any integer; we shall show that $\mathscr{B}$ satisfies the conditions of the lemma. For each $x \in X$ there is an integer $n$ such that $x \in A_n$, since the $A_n$ form a partition of $X$; thus $x \in E_n$ and there exists $V \in \mathfrak{S}_n$ such that $x \in V$; so $x \in V \cap A_n$, and we have proved that $\mathscr{B}$ is a covering of $X$. Clearly, this covering refines $\mathfrak{S}$. On the other hand, for each $x \in X$ there exists an integer $n$ such that $x \in U_n$; since $U_n$ is open and the $\mathfrak{S}_m$ are locally finite families, there exists a neighbourhood $W_m$ of $x$, for each $m$, contained in $U_n$, which meets only a finite number of sets of $\mathfrak{S}_m$; hence the neighbourhood $W = \bigcap_{m=1}^n W_m$ of $x$ meets only a finite number of sets of $\mathscr{B}$, because $W \cap A_p = \varnothing$ for $p > n$. Hence $\mathscr{B}$ is locally finite, and Lemma 4 is proved.