Theorem 5

A locally compact space $X$ is paracompact if and only if $X$ is the sum of a family of locally compact $\sigma$-compact spaces.

Suppose $X$ is paracompact, and for each $x \in X$ let $V_x$ be a relatively compact open neighbourhood of $x$ in $X$. Then by hypothesis there exists a locally finite open covering $(U(\alpha))_{\alpha \in A}$ of $X$ which refines the covering $(V_x)_{x \in X}$. The $U(\alpha)$ are therefore relatively compact. Every compact subset $K$ of $X$ meets only a finite number of the sets $U(\alpha)$, for the non-empty sets $U(\alpha) \cap K$ form a locally finite open covering of the compact space $X$; hence they must be finite in number (no. 1). Now let $R$ be the following relation between two points $x, y$ of $X$: "there exists a finite sequence $(\alpha_i)_{1 \leq i \leq n}$ of indices in $A$ such that $x \in U(\alpha_1), \ y \in U(\alpha_n)$ and $U(\alpha_i)$ meets $U(\alpha_{i+1})$ for $1 \leq i \leq n - 1$".

It is immediately verified that $R$ is an equivalence relation, and that each equivalence class mod $R$ is an open subset of $X$ [since the $U(\alpha)$ are open sets]. $X$ is therefore the sum of the locally compact subspaces (no. 7, Proposition 13) formed by the equivalence classes mod $R$, and it remains to show that each of these subspaces is the union of a countable subfamily of the family $(U(\alpha))$.

Let $x$ be any point of $X$, and define a sequence $(C_n)$ of relatively compact open subsets of $X$ by induction on $n$ as follows; $C_1$ is the union of the sets $U(\alpha)$ which contain $x$, and for each $n > 1$, $C_n$ is the union of the sets $U(\alpha)$ which meet $C_{n-1}$. It is immediately verified by induction on $n$ that each of the $C_n$ is relatively compact and is the union of a finite number of sets $U(\alpha)$. Furthermore, the equivalence class of $x$ with respect to $R$ is the union of the $C_n$: for if $(\alpha_i)_{1 \leq i \leq n}$ is a sequence of indices such that $x \in U(\alpha_1)$ and $U(\alpha_i)$ meets $U(\alpha_{i+1})$ for $1 \leq i \leq n - 1$, then one sees by induction on $i$ that $U(\alpha_i) \subset C_i$ for $1 \leq i \leq n$. It follows that the equivalence classes mod $R$ are $\sigma$-compact, and this completes the proof of the first part of the theorem.

To prove the converse, we may assume (by Proposition 18) that $X$ is $\sigma$-compact. Let $\mathfrak{R} = (G_\lambda)_{\lambda \in L}$ be any open covering of $X$, and let $(U_n)$ be a sequence of relatively compact open sets in $X$ which have the properties stated in Proposition 15 of no. 9. Let $K_n$ denote the compact set $\overline{U}_n - U_{n-1}$ ($U_n = \emptyset$ if $n \leq 0$). The open set $U_{n+1} - \overline{U}_{n-2}$ is a neighbourhood of $K_n$ by construction; hence for each $x \in K_n$ there is a neighbourhood $W_x$ of $x$ contained in one of the sets $G_\lambda$ and contained also in $U_{n+1} - \overline{U}_{n-2}$. Since $K_n$ is compact, a finite number of the sets $W_x$ cover $K_n$; let $H_{ni} (1 \leq i \leq p_n)$ be these sets. Then the family $\mathfrak{R}'$ of sets $H_{ni} (n \geq 1, 1 \leq i \leq p_n \text{ for each } n)$ is an open covering of $X$ which refines $\mathfrak{R}$, and hence to complete the proof we have to show that $\mathfrak{R}'$ is locally finite. Let $z$ be any point of $X$, $n$ the smallest integer such that $z \in U_n$; then since $z \notin U_{n-1}$, there is a neighbourhood $T$ of $z$ which is contained in $U_n$ and does not meet $U_{n-2}$. It follows that $T$ meets only those sets $H_{mi}$ for which $n - 2 \leq m \leq n + 1$, i.e. $T$ meets only a finite number of sets of $\mathfrak{R}'$.

In the course of the proof we have also established the following result: