LemmaStatement: Literature-sourcedProof: AI-generatedverified 2026-09-26 (gpt-6-sol)

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Asymptoticity of Gromov sequences is an equivalence relation

Statement

In a proper geodesic hyperbolic space, asymptoticity of Gromov sequences is an equivalence relation.

Facts & Assumptions

Given: A proper geodesic hyperbolic space $X$ with basepoint $o$ .

[L1]

The slim-triangle definition of hyperbolicity gives a Gromov-product inequality $(u, w)_{o} \geq \min {(u, v)_{o}, (v, w)_{o}} - 3 δ$ for a fixed $δ \geq 0$ (Slim triangles imply the gromov product inequality).

[A1]

Reflexivity and symmetry are immediate from the definition of asymptoticity.

Proof

technique · direct

1.1A1

By [A1], every Gromov sequence is asymptotic to itself, and if $(x_{n})$ is asymptotic to $(y_{n})$ then $(y_{n})$ is asymptotic to $(x_{n})$ .

2.1L1step 1.1algebra∎

Suppose $(x_{n})$ is asymptotic to $(y_{n})$ and $(y_{n})$ is asymptotic to $(z_{n})$ . By [L1], there is $δ \geq 0$ with $(x_{m}, z_{n})_{o} \geq \min {(x_{m}, y_{k})_{o}, (y_{k}, z_{n})_{o}} - 3 δ$ for all $m, n, k$ . Given $R > 0$ , choose $N$ so both mixed products exceed $R + 3 δ$ whenever both of their indices are at least $N$ . For every $m, n \geq N$ , taking $k = N$ in the inequality gives $(x_{m}, z_{n})_{o} > R$ . Hence $(x_{m}, z_{n})_{o} \to \infty$ , so $(x_{n})$ is asymptotic to $(z_{n})$ .

Depends on

Used by

FALSE: a proposed Gromov boundary quotient needs no equivalence check False statement

Cited to discharge well-definedness by The Gromov boundary via asymptotic sequences.

Dependency tree · two levels

5 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

Brian H. Bowditch, A course on geometric group theory, Section 5.3 (standard reference, not scraped)