Proof. For each $E\in\cali{E},\curl{F\in\cali{E}:F\cap\cl{E}\neq\emp}$ is finite by Lemma 3.12.2. Let

then $F_{E}\in\cali{N}(\cl{E})$ is precompact and open. Let $N\suf X$ and $E\in\cali{E}$. If $N\cap F_{E}\neq\emp$, then there exists $F\in\cali{E}$ such that $N\cap F\neq\emp$ and $F\cap\cl{E}\neq\emp$. Thus

By Lemma 3.12.3, $\curl{\cl{E}:E\in\cali{E}}$ is also locally finite. Hence for each $x\in X$, there exists $N\in\cali{N}(x)$ such that $\curl{F\in\cali{E}:N\cap F\neq\emp}$ is finite so $\curl{F_E}_{E\in\cali{E}}$ is locally finite.

For each $x\in X$, there exists $E\in\cali{E}$ and $N_{x}\in\cali{N}^{o}(x)$ precompact such that $x\in N_{x}\suf\cl{N_x}\suf E$. By compactness of $\cl{E}$ there exists $X_{E}\suf X$ finite such that

1. (a)

   $\curl{N_x:x\in X_E}$ is a covering of $\cl{E}$

2. (b)

   For every $x\in X_{E},N_{x}\cap E\neq\emp$

Let $X_{\cali{E}}=\cups{}{E\in\cali{E}}X_{E}$ and for each $E\in\cali{E}$ let

then $\curl{G_E}_{E\in\cali{E}}$ is an open cover of $X$. Since $G_{E}\suf E$ for all $E\in\cali{E},\curl{G_E }_{E\in\cali{E}}$ is locally finite. Let $F\in\cali{E},x\in X_{F}$ then $N_{x}\cap F\neq\emp$. Fix $E\in\cali{E}$, if $N_{x}\suf E$ then $E\cap F\neq\emp$ thus

is finite by Lemma 3.12.2, so

$\square$