Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 133 / 4

Let $[u,v]$ be one side of the geodesic quadrilateral, and draw a diagonal from $u$ to the opposite vertex. A point $p\in [u,v]$ lies, by $\delta$-thinness of the first geodesic triangle, within $\delta$ either of an adjacent side or of the diagonal. In the latter case, $\delta$-thinness of the second triangle places the nearby point of the diagonal within another $\delta$ of one of the other two sides. The triangle inequality then places $p$ within $2\delta$ of the remaining three sides. The argument applies to every side, proving the geodesic quadrilateral in a hyperbolic metric space bound.

Let $p$ be the midpoint of a geodesic $[y_1,y_2]$. Since $Y$ is a convex subset of a geodesic metric space, $p\in Y$. In the geodesic triangle with vertices $x,y_1,y_2$, the point $p$ lies within $\delta$ of $[x,y_1]$ or $[x,y_2]$. By symmetry suppose $q\in [x,y_1]$ and $d(p,q)\le \delta$. Put $L=d(y_1,y_2)$. Thenand consequentlyBecause $y_1$ is a closest point of $Y$ to $x$ and $p\in Y$, $d(x,y_1)\le d(x,p)$. Combining the inequalities gives $L\le 4\delta$. This is the coarse uniqueness of a closest point in a hyperbolic metric space.

Let $p$ be the midpoint of a geodesic $[y_1,y_2]\subseteq Y$ and put $L=d(y_1,y_2)$. Apply the geodesic quadrilateral in a hyperbolic metric space bound to the quadrilateral with consecutive vertices $y_1,z_1,z_2,y_2$. The point $p$ is within $2\delta$ of one of the other three sides. It cannot be within $2\delta$ of $[z_1,z_2]\subseteq Z$, because every point of $Y$ has distance greater than $2\delta$ from every point of $Z$.

By symmetry there is therefore a point $q\in [y_1,z_1]$ with $d(p,q)\le 2\delta$. The triangle inequality givesand henceSince $p\in Y$ and $y_1$ is a closest point of $Y$ to $z_1$, we also have $d(z_1,y_1)\le d(z_1,p)$. Therefore $L\le 8\delta$, as required.