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erdos_unit_distance

Disproved

by tianyipeng · Jun 1, 2026 · Mathlib 0df444a (Lean v4.33.1)

combinatoricsgeometry

⚠️ Retired — specification defect

The Lean statement below does not encode the problem shown on this page, so its Disproved status carries no information about that problem. Do not import this node or use it as a dependency.

Erdős unit distance problem: At most how many unit distances can appear among n points in the plane? Best known: O(n^{4/3}). Erdős conjectured O(n^{1+ε}) for all ε > 0. Best lower bound: Ω(n^{1+c/log log n}). Huge gap between upper and lower bounds.

Why this node was retired

The posted statement is

import Mathlib

theorem erdos_unit_distance :
    ∀ eps : ℝ, 0 < eps →
    ∃ C : ℝ, 0 < C ∧
    ∀ (n : ℕ) (pts : Fin n → ℝ × ℝ),
      Function.Injective pts →
      {p : Fin n × Fin n | p.1 ≠ p.2 ∧
        (pts p.1).1 ^ 2 + (pts p.1).2 ^ 2 = 1}.ncard ≤ C * n ^ (1 + eps) := by
  sorry

The counted predicate checks xᵢ²+yᵢ²=1 and ignores the second point; all distinct pairs of points on the unit circle are counted.

The recorded counterexample refutes the statement as encoded. It says nothing about the problem shown above, which is a different proposition.

Proposed corrected statement

Count unordered distinct pairs {i,j} satisfying (xᵢ−xⱼ)²+(yᵢ−yⱼ)²=1, with the intended Euclidean injective configuration and asymptotic exponent.

Diagnosis and correction from the public Prove2Me statement audit (wamlat/prove2me-errors). The correction is natural-language mathematics and is not Lean-verified — it is a specification for a corrected node, not a drop-in replacement. No corrected replacement node exists yet.

Preamble
import Mathlib
Formal statement
import Mathlib

theorem erdos_unit_distance :
    ∀ eps : ℝ, 0 < eps →
    ∃ C : ℝ, 0 < C ∧
    ∀ (n : ℕ) (pts : Fin n → ℝ × ℝ),
      Function.Injective pts →
      {p : Fin n × Fin n | p.1 ≠ p.2 ∧
        (pts p.1).1 ^ 2 + (pts p.1).2 ^ 2 = 1}.ncard ≤ C * n ^ (1 + eps) := by
  sorry
Source
https://en.wikipedia.org/wiki/Unit_distance_problem

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