Proof. $(\bracs{F_E}_{E \in \ce})$: For each $E \in \ce$, $\bracs{F \in \ce|F \cap \ol E \ne \emptyset}$ is finite by Lemma 4.18.2. Let

then $F_{E} \in \cn(\ol{E})$ is precompact.

Let $N \subset X$ and $E \in \ce$. If $N \cap F_{E} \ne \emptyset$, then there exists $F \in \ce$ such that $N \cap F \ne \emptyset$ and $F \cap \ol{E}\ne \emptyset$. Thus

By Lemma 4.18.3, $\bracsn{\ol E|E \in \ce}$ is also locally finite. Hence for every $F \in \ce$, $\bracsn{E \in \ce|F \cap \ol{E} \ne \emptyset}$ is finite.

Let $x \in X$, then there exists $N \in \cn(x)$ such that $\bracs{F \in \ce|N \cap F \ne \emptyset}$ is finite. In which case, $\bracs{E \in \ce|N \cap F_E \ne \emptyset}$ is finite as well. Therefore $\bracs{F_E}_{E \in \ce}$ is locally finite.

$(\bracs{G_E}_{E \in \ce})$: For each $x \in X$, there exists $E \in \ce$ and $N_{x} \in \cn^{o}(x)$ precompact with $x \in N_{x} \subset \ol{N_x}\subset E$.

For any $E \in \ce$, $\ol{E}$ is compact, so there exists $X_{E} \subset X$ finite such that

1. $\ol{E}\subset \bigcup_{x \in X_E}N_{x}$.

2. For every $x \in X_{E}$, $N_{x} \cap E \ne \emptyset$.

Let $X_{\ce}= \bigcup_{E \in \ce}X_{\ce}$, and for each $E \in \ce$, let

then $\bracs{G_E}_{E \in \ce}$ is an open cover of $X$. Since $G_{E} \subset E$ for all $E \in \ce$, $\bracs{G_E}_{E \in \ce}$ is locally finite.

It remains to show that $\ol{G_E}\subset E$. Let $x \in X_{F}$ such that $N_{x} \subset E$, then $N_{x} \cap F \ne \emptyset$. Since $N_{x} \subset E$, $E \cap F \ne \emptyset$. Thus

is finite by Lemma 4.18.2, so

by Proposition 4.5.4.$\square$