Distance between geodesic rays 2 This is a follow up to my previous question which turned out to be wrong. Now my question instead is this:
Given a $\delta$-hyperbolic space $(X,d)$ (with Rips triangle condition: a point on any side is contained in
the $\delta$-neighbourhood of the union of the other two sides) and two (unit speed) geodesic rays $\gamma_1, \gamma_2$ originating from the same point $x$. If there is some $t$ such that
$d(\gamma_1(t), \gamma_2(t)) \leq \delta$ then for any $0 \leq t_0 \leq t$ there exists a $t_1$ such that $d(\gamma_1(t_0), \gamma_2(t_1)) \leq \delta$.
In particular if we found a point in a geodesic triangle which connects (with distance less than $\delta$) to one side, then the points closer to the starting vertex also prefer to connect to the same side.
Any hints or a proof?
 A: Sadly this is still not true. Consider again the Cayley graph of $\mathbb{Z}$ with generators $\{1,-1\}$. This looks like a chain of beeds, where each bead is a circle of diameter $2$. From the circle being diameter $2$ we get that this is a $\delta = \frac{1}{2}$-hyperbolic space.
Let $\gamma_1$ be the geodesic ray that goes to $+\infty$ by always taking the upper halfcircles and let $\gamma_2$ be the geodesic ray that goes to $+\infty$ by taking the lower halfcircles.
Consider $t= \frac{3}{4}$. Then we have that $d(\gamma_1(t),\gamma_2(t))=\delta$. Now choose $ \frac{1}{2} < t_0 < \frac{3}{4}$ and notice that the closest point on $\gamma_2$ is more than $\delta$ from $\gamma_1(t_0)$ away.
Here is what i can prove:
Lemma:
Let $X$ be a $\delta$-hyperbolic space and $\gamma_1, \gamma_2$ two unit speed geodesic rays starting from the same point $x$. Let $0 \leq t$ be so that $d(\gamma_1(t),\gamma_2(t)) \leq \delta$. Let $0 \leq t_0 \leq t_0$. Then there exists a $0 \leq t_1 \leq t$ such that $d(\gamma_1(t_0) , \gamma_2(t_1)\leq 2 \delta$.
Proof:
Consider any triangle formed by the points $x,\gamma_1(t)$ and $\gamma_2(t)$. We apply the triangle condition to the point $\gamma_1(t_0)$ on the side $\overline{x \gamma_1(t)}$. This point has to be included in the $\delta$ neighborhood of the other two sides. If it is included in the $\delta$-neighborhood of $\overline{x,\gamma_2(t)}$, then we find $t_1$ with the required property (even with $\leq \delta$). If the point is included in the $\delta$ neighborhood of the other side $\overline{\gamma_1(t) \gamma_2(t)}$, then let $p\in \overline{\gamma_1(t) \gamma_2(t)}$ be a point with $d(\gamma_1(t),p)\leq \delta$. We set $t_1=t$ and use the triangle inequality to show $d(\gamma_1(t_0),\gamma_2(t_1))\leq d(\gamma_1(t_0),p) + d(p, \gamma_2(t_1)) \leq 2\delta$. $\quad \square$
By symmetrizing this argument, you can even get a bound of $\frac{3}{2}\delta$, which is the best possible due to the above example.
