I'm trying to study Hyperbolic geometry, but I can not understand the following statement.

Let $X$ be a $δ$-hyperbolic space. Then, there exists $M > 0$ such that for any geodesic $γ$, and any ball $B_r(x)$ of arbitrary radius that is disjoint from $γ$, the closest-point projection of the ball to the geodesic has diameter $≤ M$.

I want to know the proof of the question. (Maybe $M = 5δ$)

If anyone could help, It would be really appreciated.Thanks.


Hint: Notice that the statement is easy to prove when $X$ is a real tree. Therefore, suppose the general statement fails in some hyperbolic space $X$ and take a sequence of balls $B_n$ and geodesics $\gamma_n$ such that the projection of $B_n$ on $\gamma_n$ has diameter at least $n$; then, deduce that your statement fails in some ultralimit $\left( \frac{1}{n} X , y_n \right) \to (Y,y)$. Because $Y$ is a real tree, you get a contradiction.

EDIT: In Géométrie et théorie des groupes, les groupes hyperboliques de Gromov (proposition 10.2.1 page 108), a more general statement is proved using a quantitative approximation of finite subspaces by simplicial trees. In particular, if $B_r$ is ball at distance at least $r$ from a geodesic $\gamma$, then the projection of $B_r$ on $\gamma$ has diameter at most $12 \delta$.

  • $\begingroup$ Thank you for answering. I can understand that the statement is true if X is a real tree. But, what's n and y_n ? Is n a radius of B ? $\endgroup$ – Tai Nov 23 '14 at 11:04
  • $\begingroup$ And why the lim X(δ-hyperbolic)/n = Y(a real tree)? $\endgroup$ – Tai Nov 23 '14 at 11:10
  • $\begingroup$ $n$ is just the index of the sequence; it corresponds also to a lower bound on the diameter of the projection of $B_n$ on $\gamma_n$, by definition. Then, $Y$ is a real tree because it is a $0$-hyperbolic geodesic space. Of course, you have to choose $y_n$ carefully, but I just gave a hint. $\endgroup$ – Seirios Nov 23 '14 at 13:09
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    $\begingroup$ I found a more quantitative solution in the book Géométrie et théorie des groupes, les groupes hyperboliques de Gromov. $\endgroup$ – Seirios Nov 24 '14 at 8:29
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    $\begingroup$ Yes, it is the main idea. But the conclusion does not immediately follow, further work is needed. $\endgroup$ – Seirios Feb 4 '15 at 19:57

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