Gowers' Reimer-conditions abundance conjecture

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Let [n]={1,…,n}[n]=\{1,\ldots,n\} and let S⊆P([n])\mathcal{S}\subseteq\mathcal{P}([n]). Suppose that S\mathcal{S} satisfies Reimer's conditions: there is a filter F⊆P([n])\mathcal{F}\subseteq\mathcal{P}([n]) and a bijection A↦FAA\mapsto F_A from S\mathcal{S} to F\mathcal{F} such that

A⊆FAA\subseteq F_A

for every A∈SA\in\mathcal{S}, and

[A,FA]∩[B,FB]=∅[A,F_A]\cap[B,F_B]=\emptyset

whenever A,B∈SA,B\in\mathcal{S} are distinct. Here [A,B]={C:A⊆C⊆B}[A,B]=\{C:A\subseteq C\subseteq B\}, and the abundance condition means that some element belongs to at least half of the sets. Gowers' conjecture. Any family that satisfies Reimer's conditions satisfies the abundance condition. Raz disproved this conjecture by constructing a counterexample, so the strengthening is refuted even though the original union-closed sets conjecture remains open.

References

Primary source

Kengbo Lu and Abigail Raz, “Note on the union-closed sets conjecture and Reimer's average set size theorem”, arXiv:2405.10639 (2024).

Progress summary

Refreshed
Claimed progress

A counterexample reported by Raz refutes the strengthening, and later work claims infinitely many more examples, while the original union-closed sets conjecture remains open.

The conjecture, posed in the 2016 polymath project and attributed there to Balla, asserts that Reimer’s conditions force some element to occur in at least half the sets. Raz’s 2017 note reports a counterexample, so the strengthening is false if that construction is correct.

Known results

  • Raz, 2017: a counterexample has n=8n=8 and ∣S∣=11|\mathcal{S}|=11.
  • Raz, 2017: no counterexample exists with n<8n<8 or ∣S∣<11|\mathcal{S}|<11.
  • The original union-closed sets conjecture remains open.

May 30, 2024 infinite construction

Lu and Raz’s version-2 paper claims an infinite family of counterexamples, including examples with arbitrarily large prescribed lower bounds on member-set sizes; for x≥2x\ge 2, it uses n=4x+4n=4x+4. The paper also proves the abundance conclusion under the additional assumption that every set in the associated filter has size at least n−1n-1. These claims have not been independently verified here.

Current status (as of September 2026): Raz’s counterexample and Lu–Raz’s infinite family are reported but unverified in this record; the strengthening is therefore treated as claimed solved, while the original union-closed sets conjecture remains open.

Sources

Solutions 0

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