Group acting properly discontinuous but not cocompactly on a metric space

Solution 1:

The intuition you expressed in your comment is spot on: you can do this by imitating certain steps from the proof of the Milnor-Svarc lemma.

To start, assume $D(x_0,r)$ is connected. Consider the corresponding union of closed balls $$\overline D(x_0,r) = \bigcup_{\gamma \in \Gamma} \gamma \cdot \bar B_r(x_0) $$ Using proper discontinuity, the set of those $\gamma$ for which $\bar B_r(x_0) \cap \gamma \cdot \bar B_r(x_0) \ne \emptyset$ is finite; let that set be $S$. This finite set $S$ generates the group $\Gamma$. To see why, pick any $\gamma \in \Gamma$. Using connectivity of $\bar D(x_0,r)$ there exists a sequence $\text{Id}=\delta_0,\delta_1,\ldots,\delta_M = \gamma \in \Gamma$ such that $\delta_{k-1} \cdot \bar B_r(x_0) \cap \delta_k \cdot \bar B_r(x_0) \ne \emptyset$. It follows that $\delta_{k-1}^{-1} \delta_k \in S$ for all $k$, and we have $$\gamma = (\delta_0^{-1} \delta_1) (\delta_1^{-1}\delta_2) \cdots (\delta_{K-1}^{-1} \delta_k) $$

For the other direction, pick a finite generating set $S \subset \Gamma$ and map the Cayley graph $C(\Gamma;S)$ to $X$ by a $\Gamma$-equivariant continuous function: the base point $p \in C(\Gamma;S)$ goes to $x_0$; every other vertex $\delta \cdot p$ goes to $\delta \cdot x_0$; and for each generator $\gamma \in S$ and each $\delta \in \Gamma$ the edge $\overline{\delta p, \delta\gamma p}$ goes to a geodesic $\overline{\delta \cdot x_0, \delta \gamma \cdot x_0}$. For each $\gamma \in S$, as $\delta \in \Gamma$ varies the length is independent of $\delta$. These lengths are therefore represented by the lengths of the finite set of geodesics $\overline{x_0, \gamma \cdot x_0}$ for $\gamma \in S$, so the lengths are strictly bounded above by some number $r$. Now verify, using that value of $r$, that the set $D(x_0,r)$ is connected: the Cayley graph is connected; its continuous image in $X$ is therefore connected, and is clearly a subset of $D(x_0,r)$; and each individual metric ball $D(x_0,r)$ is connected and contains a point of the Cayley graph.