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Retain the $(\lambda ,\epsilon )$-quasi-isometry $f:X\to T$ and the Morse lemma for quasi-geodesics constant $R$. Given $x,y\in X$, let $m$ be the midpoint of a geodesic $[x,y]$. There is a point $z$ on the tree geodesic $[f(x),f(y)]$ with

For any continuous path $\alpha$ from $x$ to $y$, choose a partition fine enough that consecutive $\alpha (t_i)$ are at distance at most one. Consecutive images under $f$ are then at distance at mostRemoving $z$ separates $f(x)$ from $f(y)$ in the tree along their geodesic, so this finite $D$-chain must contain an element within $D$ of $z$. For the corresponding $t_i$,The lower quasi-isometry inequality yieldsThis constant depends only on the chosen quasi-isometry, so every quasi-tree has the bottleneck property.

Solved by gpt-5.6-sol high.

The hyperbolic plane ${\mathbb{H}}^2$ is a geodesic Gromov-hyperbolic metric space for some universal constant ${\delta }_0$. It is not a quasi-tree. Indeed, for every $C>0$, choose two points on opposite sides of a large closed metric ball centred at the midpoint of their joining geodesic. The complement of that ball in ${\mathbb{H}}^2$ is path connected, so the endpoints can be joined by a continuous path that stays more than $C$ from the midpoint. Thus ${\mathbb{H}}^2$ fails the bottleneck property, whereas part (b) shows that every quasi-tree satisfies it.

Solved by gpt-5.6-sol high.