Asymptotic Cones and the Sublinear Triangle Criterion

We first discharge the reciprocal input underlying the real-field/completeness interface. If $a\in \mathcal{C}\setminus \mathcal{N}$, choose rational $\delta >0$ and $N_0\ge 1$ with $\mathrm{\mid }{a}_n\mathrm{\mid }>\delta$ for $n\ge N_0$. Set $b_n=1$ for $n<N_0$ and $b_n=1\mathrm{/}{a}_n$ otherwise; all denominators used are nonzero.

The sets containing a tail of $\mathbb{N}$ form a proper filter: intersections contain the later tail, supersets preserve containment, and no tail is empty. AC and [F8] extend it to an ultrafilter. No finite set belongs to the extension because a disjoint tail belongs; in particular it is not principal.

Least-upper-bound completeness implies that the natural multiples of $1$ are unbounded: if their supremum were $s$, then $s-1$ would not be an upper bound, giving a natural $k>s-1$ and $k+1>s$. Consequently $(B-A)2^{-j}\to 0$ for $A\le B$, since $2^j\ge j+1$.

For a sequence $u_n\in [A,B]$, repeatedly bisect the current interval $I_j=[a_j,b_j]$, retaining its left half if that half's preimage under $u$ is large, otherwise its right half. The right half is large in the latter case: intersect the large preimage of $I_j$ with the large complement of the left-half preimage. Thus each $I_j$ has large preimage, the intervals nest, and $b_j-a_j=(B-A)2^{-j}$. This is a deterministic recursion.

For rational $\epsilon >0$, take $N\ge N_0$ also beyond a Cauchy index of $a$ at tolerance $\epsilon {\delta }^2$. For $m,n\ge N$, $\mathrm{\mid }{b}_m-b_n\mathrm{\mid }=\mathrm{\mid }{a}_m-a_n\mathrm{\mid }\mathrm{/}(\mathrm{\mid }{a}_m\mathrm{\mid }\mathrm{\mid }{a}_n\mathrm{\mid })\le \mathrm{\mid }{a}_m-a_n\mathrm{\mid }\mathrm{/}{\delta }^2<\epsilon$. The nonstrict comparison includes $a_m=a_n$. Thus $b\in \mathcal{C}$.

Set $u={\sup }_j{a}_j$. All $a_j\le u\le b_j$: for fixed $j$, every later left endpoint lies below $b_j$, and earlier ones lie below $a_j$. For any $\epsilon >0$, a sufficiently late $I_j$ has length less than $\epsilon$, so its large preimage lies in $\{n:\mathrm{\mid }{u}_n-u\mathrm{\mid }<\epsilon \}$. Thus $u$ is an ultralimit, also when $A=B$.

The sequence $ab-1$ is eventually zero, hence null. The ideal $\mathcal{N}$ is proper since the constant $1$ fails the null test with tolerance $1\mathrm{/}2$. Every ideal containing $\mathcal{N}$ and $a$ contains $ab-(ab-1)=1$, hence all of $\mathcal{C}$. This proves its maximality and the reciprocal interface directly, without the erroneous strict intermediate comparison in the older maximality proof. The real completeness conclusion of [F7] is used with this corrected input.

If $u\mathrm{\ne }{u}^{\mathrm{\prime }}$, their neighbourhoods of radius $\mathrm{\mid }u-u^{\mathrm{\prime }}\mathrm{\mid }\mathrm{/}3$ are disjoint by the triangle inequality. Their preimages cannot both belong to a proper filter. Hence the limit is unique. Equality of two sequences on a large set lets every neighbourhood test for one pass to the other by intersection; their limits therefore agree. A free ultrafilter contains no finite set: if a finite set were large but none of its singletons were large, intersecting their large complements from [F9] would contradict its largeness. A large singleton would make the ultrafilter principal by upward closure and the filter intersection axiom. Hence the complement of every finite set is large by [F9], so an ordinary convergent bounded sequence satisfies the same tests.

Intersect the two large error sets for $u_n,u$ and $v_n,v$, each at tolerance $\epsilon \mathrm{/}2$. There $\mathrm{\mid }(u_n+v_n)-(u+v)\mathrm{\mid }\le \mathrm{\mid }{u}_n-u\mathrm{\mid }+\mathrm{\mid }{v}_n-v\mathrm{\mid }<\epsilon$. For $c\mathrm{\ne }0$ use tolerance $\epsilon \mathrm{/}\mathrm{\mid }c\mathrm{\mid }$ to obtain $\mathrm{\mid }c{u}_n-cu\mathrm{\mid }<\epsilon$; for $c=0$ the sequence is constant zero. Uniqueness identifies the stated limits.

If $\mathrm{\mid }{u}_n\mathrm{\mid }\le H$, intersect the large sets where both errors are less than $\epsilon \mathrm{/}(H+\mathrm{\mid }v\mathrm{\mid }+1)$. Then $\mathrm{\mid }{u}_n{v}_n-uv\mathrm{\mid }\le H\mathrm{\mid }{v}_n-v\mathrm{\mid }+\mathrm{\mid }v\mathrm{\mid }\mathrm{\mid }{u}_n-u\mathrm{\mid }<\epsilon$. Also $\mid \mathrm{\mid }{u}_n\mathrm{\mid }-\mathrm{\mid }u\mathrm{\mid }\mid \le \mathrm{\mid }{u}_n-u\mathrm{\mid }$, so absolute values converge to $\mathrm{\mid }u\mathrm{\mid }$. These arguments apply to zero and unit constants as well.

If $u_n\le v_n$ on a large set but $u>v$, intersect that set with the two error sets of radius $(u-v)\mathrm{/}3$. It would give $u_n>v_n$, impossible. Thus $u\le v$. All calculus constructions used only deterministic interval bisection and finite intersections once the ultrafilter was supplied; AC was spent only in the free-filter existence step.

Symmetry and the triangle inequality give $0=d_n(x_n,x_n)\le 2d_n(x_n,y_n)$, so distances are nonnegative. The bound ${\lambda }_n{d}_n(x_n,y_n)\le {\lambda }_n{d}_n(x_n,e_n)+{\lambda }_n{d}_n(y_n,e_n)$ makes every pairwise distance sequence bounded. Thus $D(x,y)$ exists and is finite.

Passing the coordinate identities and inequalities to limits gives $D(x,x)=0$, $D(x,y)=D(y,x)\ge 0$ and $D(x,z)\le D(x,y)+D(y,z)$. In particular $D(x,y)=D(y,z)=0$ implies $D(x,z)=0$, establishing transitivity as well as reflexivity and symmetry of the zero relation.

The coordinate triangle inequalities in both orders imply $\mathrm{\mid }{\lambda }_n{d}_n(x_n,y_n)-{\lambda }_n{d}_n(x_n^{\mathrm{\prime }},y_n^{\mathrm{\prime }})\mathrm{\mid }\le {\lambda }_n{d}_n(x_n,x_n^{\mathrm{\prime }})+{\lambda }_n{d}_n(y_n,y_n^{\mathrm{\prime }})$. When $x\sim x^{\mathrm{\prime }}$ and $y\sim y^{\mathrm{\prime }}$, the right side has limit zero. Therefore $D(x,y)=D(x^{\mathrm{\prime }},y^{\mathrm{\prime }})$ and the quotient formula is independent of representatives.

On the quotient, symmetry and the triangle inequality follow from those for $D$. Distance zero means exactly $x\sim y$, which means $[x]=[y]$; conversely equal classes have distance zero. If representatives agree on a large set, their distance sequence is zero there, hence has limit zero.

If ${\lambda }_n{d}_n(x_n,e_n)\le R$ only on a large set $A$, replace $x_n$ by $e_n$ outside $A$. The replacement is globally bounded by $R$. Two such choices agree on the intersection of their large sets, so step 4.1 identifies them. Every globally bounded sequence is already of this form, proving equality of the quotient constructions, including the constant basepoint and one-point quotient.

If the bounded-neighbourhood hypothesis holds only on a large set $A$, replace $S_n$ by $\{e_n\}$ for $n\mathrm{\notin }A$. The represented limit is unchanged: intersect a representative's membership set with $A$ and replace its other coordinates by $e_n$; F1 identifies the old and new classes. The new family meets the same rescaled ball for every $n$. By AC choose $o_n\in S_n$ with ${\lambda }_n{d}_n(o_n,e_n)\le R$, and orient the given distance parameter around $o_n$. Put $a_n={\lambda }_n{\alpha }_n$, $b_n={\lambda }_n{\beta }_n$. The bounded sequence $a_n\mathrm{/}(1+a_n)$ has limit $q\in [0,1]$. If $q<1$, define $a=q\mathrm{/}(1-q)$; continuity of this rational function near $q$ gives the finite extended ultralimit. If $q=1$, then for each finite $H$, $a_n>H$ on a large set, since $a_n\le H$ implies $a_n\mathrm{/}(1+a_n)\le H\mathrm{/}(1+H)<1$. Define $a=+\mathrm{\infty }$ in this case, and define $b$ identically.

For $t\in J=[-a,b]\cap \mathbb{R}$, set $c_n(t)=\max (-a_n,\min (t,b_n))$ and $x_n(t)={\gamma }_n(c_n(t)\mathrm{/}{\lambda }_n)$. Clamping is in rescaled units; the argument of ${\gamma }_n$ is in original units. Since $0\in [-a_n,b_n]$, $\mathrm{\mid }{c}_n(t)\mathrm{\mid }\le \mathrm{\mid }t\mathrm{\mid }$ and ${\lambda }_n{d}_n(x_n(t),e_n)\le \mathrm{\mid }t\mathrm{\mid }+R$, so $x(t)$ is admissible. The finite endpoint limits, or the eventual bound at any fixed finite $t$ for an infinite endpoint, show $c_n(t){\to }_{\omega }t$, including $t=-a$ or $t=b$ when finite.

The exact identity ${\lambda }_n{d}_n(x_n(s),x_n(t))=\mathrm{\mid }{c}_n(s)-c_n(t)\mathrm{\mid }$ and real limit calculus give $d_{\omega }([x(s)],[x(t)])=\mathrm{\mid }s-t\mathrm{\mid }$. Thus $t\mapsto [x(t)]$ is an isometric embedding of $J$.

Conversely let $z_n={\gamma }_n(u_n)$ be admissible points on $S_n$ and set $t_n={\lambda }_n{u}_n$. Then $\mathrm{\mid }{t}_n\mathrm{\mid }={\lambda }_n{d}_n(z_n,o_n)\le {\lambda }_n{d}_n(z_n,e_n)+R$ is bounded, so $t={\lim }_{\omega }{t}_n$ exists. From $-a_n\le t_n\le b_n$ follows $t\in J$, including the finite endpoint inequalities. Projection onto an interval satisfies $\mathrm{\mid }{t}_n-c_n(t)\mathrm{\mid }\le \mathrm{\mid }{t}_n-t\mathrm{\mid }$ when $t_n$ lies in that interval (check $t$ below, in, or above it). Hence ${\lambda }_n{d}_n(z_n,x_n(t)){\to }_{\omega }0$ and $[z]=[x(t)]$.

If $a,b$ are finite, translate $J$ to $[0,a+b]$; if only one is infinite, translate and possibly reverse it to $[0,\mathrm{\infty })$; if both are infinite, $J=\mathbb{R}$. These are exactly the asserted interval, ray and line possibilities. When $a=b=0$, the image is one point.

For any two ultralimit classes choose representative sequences $v_n,w_n$. AC chooses a segment joining each pair. Take $o_n=v_n$, so $a_n=0$ and $b_n={\lambda }_n{d}_n(v_n,w_n)$ is uniformly bounded and has limit $d_{\omega }([v],[w])$. step 3.1 and step 3.2 give a segment with exactly these endpoints, also if that distance is zero. Thus the ultralimit is geodesic.

For any nonempty set $A$, the triangle inequality gives $d(x,A)\le d(x,y)+d(y,A)$ and the reverse inequality with $x,y$ exchanged, so $\mathrm{\mid }d(x,A)-d(y,A)\mathrm{\mid }\le d(x,y)$. Diameter of three points changes by at most twice the largest displacement when the three points move. Thus the parameter functions defining minsize and slimness are Lipschitz on products of the finite closed side intervals.

Here is the needed extrema argument without a countable choice of approximate minimizers. A finite box has dense grids obtained by dividing each coordinate interval into $2^j$ equal pieces. A Lipschitz function is bounded there by its value at one corner plus its Lipschitz constant times the box diameter. Its infimum exists by [F2]. Enumerate the union of the finite grids by level and within each level in lexicographic order; choose the first grid point with value less than the infimum plus $2^{-n}$. Density and continuity guarantee such a point. Successively bisect the box and keep the first subbox containing infinitely many terms, then take successive least indices from those subboxes. Supremums of the nested coordinate left endpoints give a limit in the box, since the widths tend to zero (the supremum argument for the unboundedness of natural numbers gives $2^{-j}\to 0$). Lipschitz continuity makes its value the infimum. Apply the same argument to the negative function for a maximum. Zero-width coordinate intervals cause no change. This proves minsize attainment and nearest-point attainment on each finite side; step 1.1 then gives slimness attainment on the finite union of sides.

Unique arcs imply unique geodesic images. For sides $[x,y]$ and $[x,z]$, every shared point $p$ has the same initial segment $[x,p]$ on both sides by arc uniqueness. Their intersection is therefore an initial interval; it is closed by compactness of the two finite segments, and is $[x,b]$ for a last point $b$. The remaining subsegments from $b$ to $y,z$ meet only at $b$. Their concatenation is an arc unless one is zero, and hence equals $[y,z]$ by uniqueness. Distances along that side add through $b$, as do distances on the other two sides. The triangle is a tripod.

Tripods are $0$-slim because each leg belongs to two sides. Conversely, in a space where all chosen triangles are $0$-slim, take two geodesics from $x$ to $y$ and the constant third side at $x$. Closedness of the other side makes each point of the first lie on it, and conversely. Their distance parameters from $x$ then agree, so geodesics are unique. Intersections of sides are closed initial segments by this uniqueness. For the last intersection $b$ of $[x,y]$ and $[x,z]$, $0$-slimness puts $[y,z]$ inside their union, and puts the portions beyond $b$ of both these sides on $[y,z]$. That side must pass through $b$ by its interval parameter; the resulting union is a tripod.

Suppose now all triangles are tripods. At a point $p$, declare $u,v\mathrm{\ne }p$ equivalent if $[p,u]$ and $[p,v]$ share a positive initial segment. Reflexivity and symmetry hold. Transitivity holds because two positive shared initial segments on $[p,v]$ have a positive common shorter segment. In a tripod, nonequivalent points satisfy $d(u,v)=d(u,p)+d(p,v)$; thus the ball of radius $d(u,p)$ about $u$ stays in its class. Every class is open in $X\setminus \{p\}$, and so is its complement. If $p$ is interior to $[x,y]$, the points $x,y$ lie in different classes.

Every continuous path from $x$ to $y$ contains each interior $p\in [x,y]$. Otherwise the inverse images $U,V$ of the class of $x$ and its complement partition $[0,1]$ into disjoint relatively open sets with $0\in U$, $1\in V$. Let $s=\sup \{t:[0,t]\subset U\}$. Openness at 0 and 1 gives $0<s<1$, and every $t<s$ belongs to $U$. If $s\in V$, a left neighbourhood contradicts that assertion; if $s\in U$, a right neighbourhood extends the initial interval past $s$. Both are impossible. This uses only [F2] and the continuity interface in [F1].

An injective path from $x$ to $y$ has no point $z$ off $[x,y]$. The tripod with vertices $x,y,z$ attaches $z$ to $[x,y]$ at a point $b$. Apply step 5.1 to the two subpaths separated by $z$: both contain $b$, using their endpoints when $b=x$ or $b=y$. Their parameter intervals meet only at $z$, and $b\mathrm{\ne }z$, contradicting injectivity. Therefore each arc equals $[x,y]$. Together with step 3.1 and step 3.2 this proves all three conditions equivalent, also for empty or one-point spaces where distinct-point assertions are vacuous.

A point $z$ has a nearest point $p$ on any segment, ray or line $S$. For a ray or line with origin $o$, the inequality $d(z,S(t))\ge \mathrm{\mid }t\mathrm{\mid }-d(z,o)$ reduces minimization to a finite closed interval; step 2.1 applies. For every $q\in S$, the tripod of $z,p,q$ has branch point $p$: any other branch point on $[p,q]$ would be closer to $z$. Thus $d(z,q)=d(z,p)+d(p,q)$. This also makes $p$ unique.

Two rays from $o$ have intersection an initial closed interval, possibly unbounded. If they split at parameter $T<\mathrm{\infty }$, the tripod distance from the first ray's point at $t>T$ to the second ray is $t-T$; hence their Hausdorff distance is infinite. If they never split, equal distance parameters give identical maps. This proves the common-origin rule.

For rays from $x$ and $y$, project $x$ to the second ray at $p$. By step 7.1, $[x,p]$ followed by its tail from $p$ is a geodesic ray from $x$ and has finite Hausdorff distance from the original second ray: the omitted initial piece and the added connector are finite. If the original two rays have finite Hausdorff distance, the new ray and the first ray do too and step 8.1 makes them coincide. Their tails beyond $p$ are therefore equal, at explicit thresholds $d(x,p)$ and $d(y,p)$.

If a line $S$ contained a point $x$ off a line $T$, let $p$ be its projection to $T$. Of the two directions of $S$ from $x$, at least one does not initially follow $[x,p]$, since these two directions intersect only at $x$. Every point at distance $t$ down that direction has its path to $T$ pass through $x,p$, by the tripod rule, and hence distance $t+d(x,p)$ from $T$. Finite Hausdorff distance excludes this. Thus $S\subset T$; an isometric copy of the full real line contained in a line is that entire line, because both parameter directions are unbounded. The two lines coincide, irrespective of orientation.

If no such $M$ exists, for each $n\ge 1$ select two sides with $q_n=d(y_n,z_n)>1$ and $D_n=d_H([x_n,y_n],[x_n,z_n])>n{q}_n$. Extrema exist on the finite sides. Exchange the endpoint names when necessary and select $a_n\in [x_n,y_n]$ attaining $d(a_n,[x_n,z_n])=D_n$. Select a nearest point $b_n$ on the other side. AC supplies these countable choices.

Base at $a_n$ and scale by $1\mathrm{/}{D}_n$. Since $D_n>n$, the scales tend to zero ordinarily, so the cone is a real tree. Both sides meet its bounded basepoint neighbourhood, at $a_n$ and $b_n$. Each point on either side has a point on the other within $D_n$. For any bounded represented sequence, such nearest points are bounded by the triangle inequality. Thus their limit images $S,T$ satisfy $d_H(S,T)\le 1$ and $d(a,T)=1$, where $a=[a_n]$: every represented point of $T$ has distance at least one from $a$, and $[b_n]$ attains one.

The rescaled distances of $y_n,z_n$ from $a_n$ differ by at most $q_n\mathrm{/}{D}_n<1\mathrm{/}n$. They therefore are both finite in the extended ultralimit or both infinite; in the finite case their classes are equal. The common endpoint $x_n$ likewise either has finite rescaled distance or escapes. Finite values can be made uniformly bounded by replacing coordinates off a large set; endpoint distance limits classify the side domains by [F1].

If both ends are finite, $S,T$ are segments with the same two endpoints and coincide by tree uniqueness. If precisely one end is finite, $S,T$ are rays with the same finite endpoint and finite Hausdorff distance, so coincide. If both ends escape, they are lines at finite Hausdorff distance, so again coincide. These cases exhaust the endpoint limits, and every case contradicts $a\in S$ and $d(a,T)=1$. Hence the asserted finite $M$ exists; enlarging it if necessary makes it positive.

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.

Choose any point $c$ on the given cone segment and a representative $c_n$. Keep the prescribed side $[a_n,b_n]$ and choose sides $[a_n,c_n]$, $[c_n,b_n]$ by AC. Their perimeters $P_n$ satisfy ${\lambda }_n{P}_n\le H$ for some finite $H$, because all three endpoint sequences are admissible and every side length is an endpoint distance.

For $\epsilon >0$ choose $P_0>0$ such that $m_X(P)\le \epsilon P$ for $P\ge P_0$. For $P<P_0$, the profile bound $m_X(P)\le P$ gives $m_X(P)\le P_0$. Thus $0\le {\lambda }_n\mathrm{minsize}({\mathrm{\Delta }}_n)\le \epsilon H+{\lambda }_n{P}_0$. The last term tends ordinarily to zero, so the scaled minsize tends to zero, with bounded and unbounded perimeters both covered.

Select a minimizing triple $x_n\in [a_n,b_n]$, $y_n\in [a_n,c_n]$, $z_n\in [c_n,b_n]$. All are admissible because the sides have bounded scaled lengths and bounded endpoints. By step 2.1 their three classes coincide at a point $x$. Exact distance additivity on the other two sides gives $d(a,c)=d(a,x)+d(x,c)$ and $d(c,b)=d(c,x)+d(x,b)$.

Since $c$ belongs to the cone segment, $d(a,b)=d(a,c)+d(c,b)=d(a,x)+2d(x,c)+d(x,b)\ge d(a,b)+2d(x,c)$. Nonnegativity forces $d(x,c)=0$, so $c=x$ lies on the prescribed side limit. This holds for every $c$ on every cone segment joining $a,b$.

The side limit and any cone segment are both isometric copies of $[0,d(a,b)]$ from $a$ to $b$. Containment from step 4.1 is equality: the point at each distance parameter $t$ on the second lies on the first and must equal its unique point at parameter $t$. If $a=b$, both intervals are a singleton. Hence the segment is unique and is precisely the prescribed limit.

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.

Rotate coordinates so distinct polygon vertices have distinct horizontal coordinates; only finitely many directions are excluded. Vertical lines through the vertices cut the complement into finitely many open-sided trapezoids, with vertical walls included except for polygon vertices. Each trapezoid is convex and therefore path connected by straight segments. Label a trapezoid left if it touches the left side of a directed polygon edge or the left sector at an extremal vertex, and right analogously. Every trapezoid receives a label: one touching an edge is labelled from that edge, and an extreme unbounded slab touches the sector of its extreme vertex.

Trace the directed polygon. Along each edge list the adjacent trapezoids on its left in the order encountered; at a horizontal-coordinate extremum insert the trapezoid in the left turning sector if necessary. Consecutive listed trapezoids share a vertical wall away from the polygon, including the first and last entries at the initial vertex. Every left-labelled trapezoid appears because it has either the defining edge contact or the defining vertex sector. Thus their union is path connected. The same traversal on the right gives a second path-connected union, and the two unions cover the complement. There are at most two path components.

For a point not on a vertical vertex line, count the edges intersecting its upward vertical ray modulo two. The parity is constant in a trapezoid and agrees across any shared wall: at a vertex the edges above a crossing wall change by zero or two if both neighbours are on the same side, and by zero if the neighbours are on opposite sides. Thus parity extends as a locally constant function throughout the complement. Above all edges it is even, and crossing one edge into an adjacent slab cell changes it to odd. Both values occur, and no continuous path can change them: the inverse images of the two values would separate its parameter interval, which is impossible as follows. If a locally constant two-valued function on $[a,b]$ had different endpoint values, let $S$ be the set of $t$ for which its value agrees with its value at $a$ throughout $[a,t]$. Local constancy at $a$ makes $S$ nonempty. Its supremum $c$ exists. Every $t<c$ lies below an element of $S$, so the value is constant on $[a,c)$. Local constancy at $c$ then gives that same value at $c$ and just beyond it if $c<b$, contradicting the supremum; if $c=b$ it contradicts the different endpoint value. Restricting a path to two points with different parities gives this contradiction Together with step 2.1 there are exactly two components. The complement outside a large rectangle is connected and even, so only the even component is unbounded. Each edge and each vertex borders both components by its local two-sector chart.