# Criterion for visuality of hyperbolic spaces

I am trying to understand the following sentence on p. 156 of Buyalo-Schroeder, Elements of asymptotic geometry: "Every cobounded, hyperbolic, proper, geodesic space is certainly visual."

Let $$X$$ be a proper geodesic hyperbolic space. $$X$$ is called cobounded if there exists $$R>0$$ such that for all $$x,y \in X$$ there exists an isometry $$f$$ of $$X$$ such that $$d(f(x), y) < R$$. It is called visual, if for some (hence any) basepoint $$o$$ of $$X$$ there exists $$R>0$$ such that every $$x\in X$$ has distance at most $$R$$ from some geodesic ray emanating from o.

From the formulation it sounds that the sentence above should be an obvious fact, but I don't see how one should approach this. Can anyone provide a hint, how one would even get started?

• Actually if $X$ is bounded, the result is technically false since there's no geodesic ray, so $X$ is not "visual" as defined here. – YCor Mar 7 at 8:02

Let $$x,y \in X$$ be two points. If $$X$$ is bounded, there is nothing to prove. Otherwise, $$X$$ must have a boundary of cardinality at least two. So we can fix two distinct points $$\zeta, \xi \in \partial X$$, and a bi-infinite geodesic $$\gamma$$ between them. As $$X$$ is cobounded, up to translating $$\gamma$$ by an isometry, we may suppose without loss of generality that $$y$$ is at a controlled distance from $$\gamma$$. Because an ideal triangle we fix $$\gamma \cup [x,\zeta) \cup [x,\xi)$$ must be thin, it is not difficult to show that $$y$$ is at a controlled distance from either $$[x,\zeta)$$ or $$[x,\xi)$$.
• You already need coboundedness to deduce that $\partial X$ has at least 2 points. – YCor Mar 7 at 8:03
• Actually I couldn't figure out if the result is false without assuming $X$ hyperbolic (only assuming $X$ unbounded, geodesic, proper, and cobounded under its isometry group). – YCor Mar 7 at 10:19
• I think the main problem is to show the existence of a single bi-infinite geodesic using coboundedness (and assuming that $X$ is unbounded). Then one can argue with thin triangles as suggested. There certainly exist unbounded proper geodesic spaces which do not contain a single bi-infinite geodesic, but I guess they cannot be cobounded. – Lyonel Mar 7 at 12:17
• The bi-infinite geodesic is not a problem. Take two sequences of points $(x_n)$ and $(y_n)$ such that $d(x_n,y_n) \to + \infty$. For every $n \geq 0$, fix a geodesic $[x_n,y_n]$. Up to translating $[x_n,y_n]$ by an isometry of $X$, suppose that the midpoint of $[x_n,y_n]$ belongs to a fixed ball. Now use the properness to find a converging subsequence of $([x_n,y_n])$. It yields a bi-infinite geodesic. The argument does not use the hyperbolicity of $X$. – AGenevois Mar 7 at 12:32