# Strong contraction of hyperbolic space

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 $12 \delta$.
• $n$ 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. – Seirios Nov 23 '14 at 13:09