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Common holes for any fixed number of subset-sum-factor maps (Erdős 624)

Proved
Erdos624.subsetSum_common_hole

by Ethan Cheung Wang · Oct 23, 2026 · Mathlib 0df444a (Lean v4.33.1)

combinatoricserdos-problems

Let UUU be a set of n=2kn=2^kn=2k elements, V=F2kV=\mathbb F_2^kV=F2k​, and consider rrr subset-sum-factor maps

gi(A)=ψi(∑a∈Avi(a)),vi:U→V, ψi:V→U arbitrary.g_i(A)=\psi_i\Bigl(\sum_{a\in A}v_i(a)\Bigr),\qquad v_i:U\to V,\ \psi_i:V\to U\ \text{arbitrary}.gi​(A)=ψi​(a∈A∑​vi​(a)),vi​:U→V, ψi​:V→U arbitrary.

Theorem. For every r≥0r\ge0r≥0 there is k0k_0k0​ such that for all k≥k0k\ge k_0k≥k0​ and all such maps g0,…,gr−1g_0,\dots,g_{r-1}g0​,…,gr−1​ there are a kkk-set Y⊆UY\subseteq UY⊆U and a point u∈Uu\in Uu∈U with

u∉⋃i<r{gi(A):A⊆Y}.u\notin\bigcup_{i<r}\{g_i(A):A\subseteq Y\}.u∈/i<r⋃​{gi​(A):A⊆Y}.

Context: Erdős Problem 624 asks whether H(n)−log⁡2n→∞H(n)-\log_2 n\to\inftyH(n)−log2​n→∞, where H(n)H(n)H(n) is the least mmm such that some f:P(X)→Xf:\mathcal P(X)\to Xf:P(X)→X (∣X∣=n|X|=n∣X∣=n) satisfies f[P(Y)]=Xf[\mathcal P(Y)]=Xf[P(Y)]=X for all ∣Y∣≥m|Y|\ge m∣Y∣≥m. By an earlier track-amplification result, it would suffice to show that for each fixed number rrr of maps gi:P(U)→Ug_i:\mathcal P(U)\to Ugi​:P(U)→U, ∣U∣=2k|U|=2^k∣U∣=2k, kkk large, some kkk-set YYY and some u∈Uu\in Uu∈U are missed by all gig_igi​ on P(Y)\mathcal P(Y)P(Y) (a common hole). This proves the fixed-rrr common-hole statement for the structured class of subset-sum-factor maps (which includes affine subset-sum maps with arbitrary relabelling of outputs). It does not cover arbitrary maps and does not settle Problem 624.

Formalization Note. U=Fin (2k)U=\mathrm{Fin}\,(2^k)U=Fin(2k), VVV is Fin k → ZMod 2, tracks indexed by Fin r\mathrm{Fin}\,rFinr.

Preamble
import Mathlib
import Definitions.Def_Erdos624_defs
open Finset
Formal statement
namespace Erdos624

theorem subsetSum_common_hole (r : ℕ) : ∃ k0, ∀ k ≥ k0,
    ∀ (v : Fin r → Fin (2 ^ k) → (Fin k → ZMod 2)) (ψ : Fin r → (Fin k → ZMod 2) → Fin (2 ^ k)),
      ∃ Y : Finset (Fin (2 ^ k)), #Y = k ∧ ∃ u, ∀ i, ∀ A ⊆ Y, ψ i (∑ a ∈ A, v 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)

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