LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10

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Sublinear minsize identifies every cone segment with a limit segment

Statement

Assume AC. Let $X$ be nonempty and geodesic with $m_{X} (P) = o (P)$ as $P \to \infty$ . In any asymptotic cone, let $a = [a_{n}]$ , $b = [b_{n}]$ . For every choice of original segments $[a_{n}, b_{n}]$ , their limit is the unique geodesic segment between $a, b$ .

Facts & Assumptions

Given: AC, a nonempty geodesic X with sublinear minsize, arbitrary basepoints and positive ordinary-null scales, endpoints a,b and an arbitrary cone segment joining them.

[F1]

The cone is geodesic and original sides have isometric represented limits. (Limits of geodesic segments, rays and lines).

[F2]

Minsize is attained for each finite triangle. (Triangle extrema and the tripod and branch rules for real trees).

[F3]

Bounded scaled distances pass sums and order to limits. (Free tail ultrafilters and bounded real ultralimit calculus).

[F4]

AC supplies countable choices of sides and minimizing triples. (The Axiom of Choice).

Proof

technique · direct

1.1

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 $λ_{n} P_{n} \leq H$ for some finite $H$ , because all three endpoint sequences are admissible and every side length is an endpoint distance.

F1F4

2.1

For $ε > 0$ choose $P_{0} > 0$ such that $m_{X} (P) \leq ε P$ for $P \geq P_{0}$ . For $P < P_{0}$ , the profile bound $m_{X} (P) \leq P$ gives $m_{X} (P) \leq P_{0}$ . Thus $0 \leq λ_{n} minsize (Δ_{n}) \leq ε H + λ_{n} P_{0}$ . The last term tends ordinarily to zero, so the scaled minsize tends to zero, with bounded and unbounded perimeters both covered.

step 1.1givenF3

3.1

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 $d (c, b) = d (c, x) + d (x, b)$ .

step 1.1step 2.1F1F2F3F4

4.1

Since $c$ belongs to the cone segment, $d (a, b) = d (a, c) + d (c, b) = d (a, x) + 2 d (x, c) + d (x, b) \geq d (a, b) + 2 d (x, c)$ . 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 $a, b$ .

step 3.1algebra

5.1

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.

step 4.1F1∎

Depends on

Used by

Cones of a line and of real trees Example
Sublinear triangle minsize implies hyperbolicity Theorem

Dependency tree · two levels

23 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Drutu–Kapovich, Geometric Group Theory — §11.21 Lemma 11.177, PDF p.449 (standard reference, not scraped)