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 $L=d(y_1,y_2)$. Then
$$
d(q,y_1)\ge d(p,y_1)-d(p,q)\ge L/2-\delta,
$$
and consequently
$$
d(x,p)\le d(x,q)+\delta
=d(x,y_1)-d(q,y_1)+\delta
\le d(x,y_1)-L/2+2\delta.
$$
Because $y_1$ is a closest point of $Y$ to $x$ and $p\in Y$, $d(x,y_1)\le d(x,p)$. Combining the inequalities gives $L\le4\delta$. This is the <coarse uniqueness of a closest point in a hyperbolic metric space>.

Solution

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

If $Y$ is convex in a $\delta$-hyperbolic metric space and two points of $Y$ both minimize distance from $x$, then their distance is at most $4\delta$. Thus a closest-point projection to $Y$, when it exists, is unique up to uniformly bounded error.


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 Incoming links: Coarse uniqueness of a closest point in a hyperbolic metric space