Proposition 18

Let $X$ be a complete metric space, let $R$ be an open equivalence relation on $X$ such that $X/R$ is Hausdorff, and let $\varphi : X \to X/R$ be the canonical mapping. Then if $K$ is any compact subset of $X/R$, there is a compact subset $K'$ of $X$ such that $\varphi(K') = K$.

Let $\mathscr{B}_1$ be the set of all open balls of radius $1/2$ in $X$. As $B$ runs through $\mathscr{B}_1$, the sets $\varphi(B)$ form an open covering of $K$, and therefore there exists a finite number of points $x_1, \ldots, x_m$ of $X$ such that the images under $\varphi$ of the open balls with radius $1/2$ and centre $x_i$ ($1 \leq i \leq m$) form an open covering of $K$. Let $H_1 = \{ x_1, \ldots, x_m \}$ and suppose that we have defined a finite set $H_i$, for $1 < i \leq n$, such that:

(i) $H_i \subset H_{i+1}$ and each point of $H_{i+1}$ is at a distance $< 1/2^i$ from $H_i$, for $1 \leq i \leq n - 1$;

(ii) the images under $\varphi$ of the open balls of radius $1/2^i$, with centres at the points of $H_i$, form an open covering of $K$, for $1 \leq i \leq n$.

Let $\mathscr{B}_{n+1}$ be the set of all open balls of radius $1/2^{n+1}$ whose centre $x$ is such that $d(x, H_n) < 1/2^n$ ($d$ being the metric on $X$). The properties of $H_n$ show that the sets $\varphi(B)$, for $B \in \mathscr{B}_{n+1}$, form an open covering of $K$; hence there is a finite set $L_{n+1} \subset X$ such that the images under $\varphi$ of the open balls of radius $1/2^{n+1}$ whose centre belongs to $L_{n+1}$ form an open covering of $K$. Taking $H_{n+1} = H_n \cup L_{n+1}$, we see that we can define inductively an infinite sequence $(H_n)$ of finite subsets of $X$ with properties (i) and (ii) above. Let $H = \bigcup_n H_n$, and let us show that $H$ is precompact. For each $p > 0$ and each point $z_{n+p} \in H_{n+p}$, there exists a sequence of points $z_{n+i} \in H_{n+i}$ ($0 \leq i \leq p - 1$) such that

it follows that $d(z_n, z_{n+p}) \leq \sum_{i=0}^{p-1} 1/2^{n+i} \leq 1/2^{n-1}$, and consequently $d(y, H_n) \leq 1/2^{n-1}$ for all $y \in H$, which proves our assertion. Since $X$ is complete, $\overline{H}$ is compact, hence $\varphi(\overline{H})$ is compact. Next, let us show that $K \subset \varphi(\overline{H})$. If $z \in K$, then by definition $d(\overline{H}_n, \overline{\varphi}(z)) \leq 1/2^n$ for all $n$, and therefore $d(\overline{H}, \overline{\varphi}(z)) = 0$; but $\overline{\varphi}(z)$ is closed and $\overline{H}$ is compact, so that this implies $\overline{H} \cap \overline{\varphi}(z) \neq \emptyset$ (no. 2, Remark following Proposition 3); hence the assertion. Thus if $K' = \overline{H} \cap \overline{\varphi}(K)$, then $K'$ is closed in $\overline{H}$ and therefore compact, and from what has been proved we have $\varphi(K') = K$.

Q.E.D.