No common hole can be forced over a prescribed -core (Erdős 624)
ProvedErdos624.prescribed_core_covercombinatoricserdos-problems
Theorem. Let , , and let be any set with . Then there are two maps such that for every -set ,
Explicitly, for a bijection . Context: Erdős Problem 624 asks whether , where is the least such that some () satisfies for all . By an earlier track-amplification result, it would suffice to show that for each fixed number of maps , , large, some -set and some are missed by all on (a common hole). This shows that the one-map palette argument over a fixed core (which proves the robust Alon bound) cannot give a two-map common hole: any successful argument must compare different cores.
Formalization Note. .
Preamble
import Mathlib import Definitions.Def_Erdos624_defs open Finset
Formal statement
namespace Erdos624
theorem prescribed_core_cover (k : ℕ) (hk : 3 ≤ k) (W : Finset (Fin (2 ^ k)))
(hW : #W = k - 2) :
∃ g : Fin 2 → Finset (Fin (2 ^ k)) → Fin (2 ^ k),
∀ Y : Finset (Fin (2 ^ k)), #Y = k → W ⊆ Y → ∀ 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; claim from GPT 6.1 Sol (fifth answer, Oct 23 2026), Lean-verified by Grok Bot for Ethan Cheung (Oct 23 2026)