Cones of a line and of real trees

If $(x_n)$ is admissible, $a_n={\lambda }_n(x_n-e_n)$ is bounded in absolute value, so $T([x_n])={\lim }_{\omega }{a}_n$ exists. For two admissible sequences, F2 gives $\mathrm{\mid }T([x_n])-T([y_n])\mathrm{\mid }={\lim }_{\omega }\mathrm{\mid }{\lambda }_n(x_n-y_n)\mathrm{\mid }=d_{\omega }([x_n],[y_n])\mathrm{.}$ Thus $T$ is constant on equivalent representatives, and equal $T$ values imply zero ultradistance and the same cone point. It is an isometric injection.

In a real tree every chosen triangle is a tripod by F4, and its branchpoint belongs to all three sides. Its minsize is therefore zero, giving the sublinear function $m_X(P)=0$. F3 makes the cone geodesic, and F5 applies under AC: each of its geodesic segments is unique and is the limit of any prescribed representative segments. This includes arbitrary moving basepoints and arbitrary positive scales tending to zero. The next branchpoint calculation establishes the additional tripod conclusion.

For $t\in \mathbb{R}$ set $x_n=e_n+t\mathrm{/}{\lambda }_n$. Then ${\lambda }_n\mathrm{\mid }{x}_n-e_n\mathrm{\mid }=\mathrm{\mid }t\mathrm{\mid }$, so this is admissible and $T([x_n])=t$. Conversely, for an admissible $(y_n)$ with $t=T([y_n])$, the distance to this inverse representative is the ultralimit of $\mathrm{\mid }{\lambda }_n(y_n-e_n)-t\mathrm{\mid }$, which is zero. Both inverse identities hold. For the concrete choice $e_n=(n+1{)}^2$, ${\lambda }_n=1\mathrm{/}(n+1)$, the sequence $x_n=(n+1{)}^2+3(n+1)$ maps to $3$, and $y_n=(n+1{)}^2-2(n+1)$ maps to $-2$; their rescaled distance is $5$ for every $n\in \mathbb{N}$, including $n=0$.

More explicitly, for three representative vertices $x_n,y_n,z_n$ choose their tree branchpoint $b_n$. It satisfies $d(b_n,x_n)\le d(x_n,y_n)$ and hence ${\lambda }_{n}d(b_n,e_n)\le {\lambda }_{n}d(x_n,e_n)+{\lambda }_{n}d(x_n,y_n)$, a bounded sequence. Thus $[b_n]$ exists. It is on each limit side, since $b_n$ belongs to each original side, and F3 parameterizes those sides by their rescaled distances from an endpoint. By step 1.2 these limit sides are the actual chosen cone sides. Write $b=[b_n]$. Each side splits at $b$ into its two endpoint legs by distance additivity. Two distinct legs toward vertices $x,y$ cannot share $q\mathrm{\ne }b$, since then $d(x,y)\le d(x,q)+d(q,y)=d(x,b)+d(b,y)-2d(b,q)<d(x,y)$. Thus they form a tripod, with zero legs permitted. Every chosen cone triangle is therefore a tripod, and F4 makes the cone a real tree. For a concrete change of isometry type, take the real tree $[0,1]$, basepoint $0$ and scales $1\mathrm{/}(n+1)$. Every representative has rescaled distance at most $1\mathrm{/}(n+1)$, so its cone is a single point, whereas the original interval has two points at distance one.