Random obstruction: fails once (Erdős 624)
ProvedErdos624.not_trackBad_randomcombinatoricserdos-problems
Let be the least such that some , , satisfies for all with (Erdős Problem 624). Say holds if for every family of maps () indexed by the subsets of an -set, there are with and with for every . By an earlier result, implies for . Theorem. If and
then is false. In particular the track method can never give more than .
From and one gets for ; then apply the counting obstruction.
Formalization Note. and the -set is , as in Erdos624.track_amplification.
Preamble
import Mathlib import Definitions.Def_Erdos624_defs open Finset
Formal statement
namespace Erdos624
theorem not_trackBad_random {n m s : ℕ} (hn : 2 ≤ n)
(h : (n : ℝ) * (Real.log n * (m + 1) + 1) ≤ 2 ^ (m + s)) :
¬ ∀ g : Finset (Fin s) → Finset (Fin n) → Fin n, ∃ Y : Finset (Fin n), #Y = m ∧
∃ u : Fin n, ∀ B : Finset (Fin s), ∀ A ∈ Y.powerset, g B A ≠ u := by sorry
end Erdos624
Source
Erdős Problem 624 (https://www.erdosproblems.com/624); definitions from Formal Conjectures 624.lean; Claim B2b of claim from GPT 6.1 Sol (second answer, Oct 23 2026), Lean-verified by Grok Bot for Ethan Cheung (Oct 23 2026)