Trivial common hole for several maps (Erdős 624)
ProvedErdos624.commonHolecombinatoricserdos-problems
Theorem. Let and be finite sets, let for , and let . If , then some is omitted by every on the subsets of :
The union has at most elements. This is the counting-range common-hole lemma for Erdős Problem 624; it only works where , so it gives no improvement over the counting bound .
Preamble
import Mathlib import Definitions.Def_Erdos624_defs open Finset
Formal statement
namespace Erdos624
theorem commonHole {U ι : Type*} [Fintype U] [DecidableEq U] [Fintype ι]
(g : ι → Finset U → U) (Y : Finset U) (h : Fintype.card ι * 2 ^ #Y < Fintype.card U) :
∃ 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 B1 of claim from GPT 6.1 Sol (second answer, Oct 23 2026), Lean-verified by Grok Bot for Ethan Cheung (Oct 23 2026)