Tree cones at all basepoints and scales imply uniform slimness

The two-side bound extends to $d_H([x,y],[x,z])\le C\max (1,d(y,z))$, with $C=4M+4$. Only $d(y,z)\le 1$ needs work. If $d(x,y)\ge 3$, let $y^{\mathrm{\prime }}$ be three units before $y$ on its side. Then $2\le d(y^{\mathrm{\prime }},z)\le 4$, so [F1] gives $d_H([x,y^{\mathrm{\prime }}],[x,z])\le 4M$; restoring the last length-three piece increases this by at most three. If $d(x,y)<3$, then $d(x,z)<4$ and both sides lie within four of their common endpoint, giving Hausdorff distance at most four.

If there is no uniform slimness bound, use AC to choose a triangle for each $n$ whose slimness $d_n>n$. Maximize distance to the other two sides over all three sides, and rename so the maximizing point is $a_n\in [x_n,y_n]$ with nearest point $b_n\in [y_n,z_n]$ at distance $d_n$. Let $c_n\in [x_n,z_n]$ be nearest to $a_n$, and put $D_n=d(a_n,c_n)\ge d_n$. Crucially every point of each of the three sides is within $d_n$ of the other two.

In the cone based at $a_n$ at scales $1\mathrm{/}{d}_n$, write $a=[a_n]$, $b=[b_n]$. It is a real tree, $d(a,b)=1$, and every represented point of either opposite side is at distance at least one from $a$. The full side $[x_n,y_n]$ survives and contains $a$; $[y_n,z_n]$ survives and contains $b$. Nearest points with bounded rescaled distances give points on the full represented limits by [F2].

First suppose the extended limit of $D_n\mathrm{/}{d}_n$ is finite. Then $c=[c_n]$ survives (modify exceptional coordinates as needed). Apply step 1.1 to the pairs of half-sides toward $x_n$, toward $y_n$, and toward $z_n$, starting respectively at $(a_n,c_n)$, $(a_n,b_n)$ and $(b_n,c_n)$. The resulting Hausdorff bounds remain finite after scaling because these three starting-point distances are bounded after scaling. Each pair limits either to segments with the same terminal endpoint or to rays with common tails by [F3]. The finite/infinite status matches in each pair because their startpoints stay at bounded distance.

It remains that $D_n\mathrm{/}{d}_n{\to }_{\omega }+\mathrm{\infty }$. Every point of the third side is then out of bounded rescaled range from $a_n$, since its distance is at least $D_n$. In particular $x_n,z_n$ escape. The two surviving sides $S,T$ are either rays from a common finite $y=[y_n]$, or lines when $y_n$ also escapes. Orient their halves toward $y$ positively, using $a$ and $b$ as origins. step 1.1 applied to $[a_n,y_n]$ and $[b_n,y_n]$ makes the two positive halves either terminate at the same $y$ or share a positive tail.

Choose a point $x^{\ast }$ common to the two terminal half-sides toward $x$: take their common finite endpoint, or a point sufficiently far down their common tail past both starting points. Choose $y^{\ast },z^{\ast }$ similarly. On the full first side the points $x^{\ast },a,y^{\ast }$ occur in that order, since its two halves have opposite signed parameters. The other full sides similarly contain $[y^{\ast },z^{\ast }]$ and $[z^{\ast },x^{\ast }]$. The finite triangle with these endpoints is a tripod in the cone; hence $a\in [x^{\ast },y^{\ast }]$ belongs to the union of its other two sides. This contradicts the distance-at-least-one conclusion of step 2.1. This construction includes finite zero-length terminal legs and does not invoke an ideal-boundary theorem.

For every fixed $t\ge 0$, take the point at rescaled parameter $t$ on $[b_n,z_n]$, clamping at its endpoint when necessary. These sequences are bounded because $d(a_n,b_n)\mathrm{/}{d}_n=1$. Global maximality in step 1.2 gives distance at most $d_n$ to $[x_n,y_n]\cup [x_n,z_n]$. The second set is farther than $D_n$ from $a_n$, so cannot supply this bound on a large set: the selected point is within $1+t$ of $a_n$ after scaling. Thus it has a bounded nearest point on the first side, and its limit has distance at most one from $S$. Every point of the entire negative half-ray of $T$ is therefore within one of $S$.

If $y$ is finite, $S,T$ are rays from $y$. Distinct rays from one origin split and their distance to each other grows without bound along either tail, by the tripod rule, contradicting step 4.2. If $y$ is infinite, the two lines share a positive tail by step 3.2. If distinct, their intersection is a closed terminal ray: it cannot have a gap by uniqueness of segments. Beyond its finite initial point their negative rays split, and the distance from a point on the negative tail of $T$ to $S$ is its distance back to that split point, which is unbounded. Again step 4.2 excludes this. Thus in both cases $S=T$, contradicting $a\in S$ and $d(a,T)\ge 1$. Both extended-ratio cases being impossible, a finite uniform slimness constant exists. The empty space has this property vacuously.