CounterexampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-26

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Quasi-isometry without bounded geometry need not preserve local ball counts

Statement refuted

If two metric spaces are quasi-isometric, then the cardinalities of their radius-one balls are uniformly comparable.

Facts & Assumptions

Given: The graph $X$ obtained from the integer line by attaching $n$ leaves to the vertex $n$ for each integer $n \geq 1$ , with every edge of length $1$ , and the usual integer line $Y = Z$ with graph metric.

[L1]

A quasi-isometry is a coarse Lipschitz map admitting a coarse Lipschitz quasi-inverse (Coarsely dense subsets, quasi-inverses and quasi-isometries).

Counterexample

technique · direct

1.1

Let $p : X \to Y$ collapse every attached leaf at $n$ to the spine vertex $n$ , and let $i : Y \to X$ be the inclusion of the spine. Both maps are $1$ -Lipschitz, $p \circ i = {id}_{Y}$ , and every vertex of $X$ lies at distance at most $1$ from $i (Y)$ . So $p$ is a quasi-isometry by [L1].

L1algebra

1.2

The radius-one ball about the spine vertex $n \geq 1$ in $X$ contains the two neighboring spine vertices, the center $n$ , and the $n$ attached leaves, so it has cardinality $n + 3$ . The radius-one ball about $n$ in $Y$ always has cardinality $3$ . These ball sizes are not uniformly comparable as $n \to \infty$ .

givenalgebra

2.1

Thus $X$ and $Y$ are quasi-isometric by step 1.1, while step 1.2 refutes the stated ball-count conclusion.

step 1.1step 1.2∎

Depends on

Coarsely dense subsets, quasi-inverses and quasi-isometries

Used by

Nothing in the library uses this result yet.

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

C. Löh, Geometric Group Theory, Sections 5.1-5.3 (standard reference, not scraped)