Two maps on points with no common hole on any 3-set (Erdős 624)
ProvedErdos624.pair_common_hole_false_k3combinatoricserdos-problems
Theorem. Let . There are two maps such that for every -set and every , some takes the value on a subset of .
So the two-map common-hole statement ('for , any two maps miss a common point on some -set'), which would give one unit of gain over the counting bound for Erdős Problem 624, is false at ; e.g. on . Each map misses half of on every 3-set, yet the missed sets are disjoint. The statement for large remains open.
Formalization Note. , maps indexed by ; special case of rank_cover.
Preamble
import Mathlib import Definitions.Def_Erdos624_defs open Finset
Formal statement
namespace Erdos624
theorem pair_common_hole_false_k3:
¬ ∀ g : Fin 2 → Finset (Fin (2 ^ 3)) → Fin (2 ^ 3),
∃ Y : Finset (Fin (2 ^ 3)), #Y = 3 ∧ ∃ u, ∀ i, ∀ A ⊆ Y, g i A ≠ u := by sorry
end Erdos624
Source
Erdős Problem 624 (https://www.erdosproblems.com/624); definitions from Formal Conjectures 624.lean; (B2) of claim from GPT 6.1 Sol (fourth answer, Oct 23 2026), Lean-verified by Grok Bot for Ethan Cheung (Oct 23 2026)