= Solution
On the unit three-ball $B^3$, the distance in the <Poincare ball model> is the infimum of the lengths
$$
L(\gamma)=\int\frac{2|\gamma'(t)|}{1-|\gamma(t)|^2}\,dt
$$
over piecewise smooth paths joining the two points. Equivalently the <hyperbolic metric> as a metric-space distance is
$$
\boxed{d(x,y)=\operatorname{arcosh}\left(1+\frac{2|x-y|^2}{(1-|x|^2)(1-|y|^2)}\right).}
$$
It has curvature-minus-one normalization; in particular $d(0,x)=2\operatorname{artanh}|x|$.
For a nonempty finite set $p_1,\ldots,p_m$, define $R(x)=\max_jd(x,p_j)$. The triangle inequality gives $|R(x)-R(y)|\leq d(x,y)$, so $R$ is continuous. Take $R_0=R(p_1)$. Any improving center satisfies $d(x,p_1)\leq R(x)\leq R_0$, and thus lies in the closed hyperbolic ball $K=\overline B(p_1,R_0)$.
This ball is compact: an <isometry> sending $p_1$ to the origin identifies it with the Euclidean closed ball of radius $\tanh(R_0/2)<1$, and the Euclidean and hyperbolic topologies agree inside $B^3$. The continuous function $R$ attains a minimum on $K$. Centers outside $K$ have $R>R_0$, so this is also the global minimum. Its associated closed ball proves \b[a minimum enclosing ball in <hyperbolic space> exists]. For the empty finite set, any center and radius zero suffice.
Back to article page
