= Solution
Retain the <metric geodesic> subdivision and orbit-density parts of the usual <Milnor–Švarc lemma> proof. Drop the step using proper discontinuity and compactness to prove that the displacement-bounded generating set is finite.
More explicitly, let $L=d_X(x,gx)$ and $m=\max\{1,\lceil L\rceil\}$. Subdivide a <metric geodesic> from $x$ to $gx$ into $m$ segments of length at most one. Choose orbit points $g_i x$ within distance $R$ of the subdivision points, choosing the endpoints exactly: $g_0=1$, $g_m=g$. Then
$$
d_X(g_{i-1}x,g_i x)\leq2R+1,
$$
so every nonidentity increment $g_{i-1}^{-1}g_i$ lies in $S$. Their telescoping product is $g$, and hence
$$
|g|_S\leq m\leq L+1.
$$
This also covers $L=0$, when the one increment may be a nontrivial stabilizer element. Conversely, for any word $g=s_1\cdots s_k$ in $S$, the <triangle inequality> and invariance under the isometric action give
$$
d_X(x,gx)\leq\sum_{i=1}^k d_X(x,s_i x)\leq(2R+1)k.
$$
Taking the shortest word gives the other inequality. Thus \b[the proof works unchanged except that the resulting generating set may be infinite]. Neither local compactness nor a finite stabilizer is used in this weaker argument.
Back to article page
