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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?

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  • $\begingroup$ Actually if $X$ is bounded, the result is technically false since there's no geodesic ray, so $X$ is not "visual" as defined here. $\endgroup$
    – YCor
    Mar 7, 2019 at 8:02

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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 $[x,\xi)$.

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  • $\begingroup$ You already need coboundedness to deduce that $\partial X$ has at least 2 points. $\endgroup$
    – YCor
    Mar 7, 2019 at 8:03
  • $\begingroup$ Sure. The properness is also implicitely used in different places. $\endgroup$
    – AGenevois
    Mar 7, 2019 at 10:08
  • $\begingroup$ 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). $\endgroup$
    – YCor
    Mar 7, 2019 at 10:19
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    $\begingroup$ 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. $\endgroup$
    – Lyonel
    Mar 7, 2019 at 12:17
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    $\begingroup$ 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 $X$. $\endgroup$
    – AGenevois
    Mar 7, 2019 at 12:32

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