Sublinear triangle minsize implies hyperbolicity

Fix a free ultrafilter, whose existence under AC follows from [F5], and arbitrary basepoints and positive ordinary-null scales. The resulting cone is geodesic and uniquely geodesic by [F1]. For any triangle in it, choose representatives of its three vertices and original sides; [F1] identifies their limits with the three specified cone sides.

These representative triangle perimeters have ${\lambda }_n{P}_n$ uniformly bounded. The estimate ${\lambda }_n{m}_X(P_n)\le \epsilon {\lambda }_n{P}_n+{\lambda }_n{P}_0$, with $P_0$ a threshold for sublinearity, shows their scaled minsize tends to zero. Minimizing triples exist by [F2], are bounded after scaling, and coalesce at a point $p$ lying on all three limit sides. AC supplies the countable family of triples.

In a uniquely geodesic space, if $p$ lies on all three sides, those sides are unions of the three legs from $p$ to the vertices. Two legs meet only at $p$: a common point $q\mathrm{\ne }p$ on the legs to $x,y$ would give $d(x,y)\le d(x,q)+d(q,y)=d(x,p)+d(p,y)-2d(p,q)<d(x,y)$. Thus the triangle is a tripod, with zero legs allowed. Every cone triangle is a tripod, hence the cone is a real tree by [F2].

The basepoints and scales were arbitrary for the fixed free ultrafilter. Apply [F3] to obtain a finite slimness constant for X. For empty X there are no chosen triangles, so the separate vacuous assertion holds without a profile.