Conjectura
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← problemsCB004openMathematics/ Combinatorics

The union-closed sets conjecture

Maintainer — open

In a family of sets closed under unions, is some element in at least half the sets?

Motivation

Péter Frankl, 1979. One of the most approachable-sounding open problems in combinatorics, and it resisted entirely for forty-three years.

In 2022 Justin Gilmer proved a constant lower bound of about 0.01 using an information-theoretic argument that surprised the field; the constant was quickly improved to about 0.38, but one half remains out of reach and the entropy method appears to have a ceiling below it.

Péter Frankl, 1979.

Lean API
import Mathlib.Data.Finset.Lattice.Basic
import Mathlib.Data.Finset.Card

namespace Conjectura.CB004

/-- Frankl's conjecture: in a finite family closed under unions and containing a
non-empty set, some element lies in at least half the sets. Justin Gilmer proved a
constant bound of about 0.01 in 2022 using an information-theoretic argument; one half
remains open. -/
def goal : Prop :=
  ∀ (F : Finset (Finset ℕ)), F.Nonempty → (∀ A ∈ F, ∀ B ∈ F, A ∪ B ∈ F) →
    (∃ A ∈ F, A.Nonempty) →
      ∃ x : ℕ, 2 * (F.filter (fun A => x ∈ A)).card ≥ F.card

end Conjectura.CB004
Definition4
FinsetFinset

structure

NonemptyF.Nonempty

abbrev

filterF.filter

def

cardF.card

theorem

Related work1
For your AI
Download as .md
I am proving a theorem in Lean 4 and submitting it to Conjectura.

## Problem CB004 — The union-closed sets conjecture

In a family of sets closed under unions, is some element in at least half the sets?

## Environment (fixed — do not assume anything newer)

- Lean toolchain: `leanprover/lean4:v4.33.0-rc1`
- Mathlib: `v4.33.0-rc1`

If a lemma you want does not exist in that Mathlib, prove it inline instead of
importing something newer.

## The exact statement I must prove

```lean
theorem solution : ∀ (F : Finset (Finset ℕ)), F.Nonempty → (∀ A ∈ F, ∀ B ∈ F, A ∪ B ∈ F) →
    (∃ A ∈ F, A.Nonempty) → ∃ x : ℕ, 2 * (F.filter (fun A => x ∈ A)).card ≥ F.card := by
  sorry
```

## The file I submit

```lean
import Conjectura.Problems.CB004.Statement

namespace Submission

theorem solution : Conjectura.CB004.goal := by
  sorry

end Submission

```

## The Lean definitions of every term in this problem

These are the actual definitions your proof will be checked against. Do not
substitute your own version of any of them.

### Finset

structure

```lean
Finset
```
Defined in Mathlib.

### Nonempty

abbrev

```lean
F.Nonempty
```
Defined in Mathlib.

### filter

def

```lean
F.filter
```
Defined in Mathlib.

### card

theorem

```lean
F.card
```
Defined in Mathlib.

## Rules — submissions violating these are rejected automatically

1. **Do not change the name or type of `solution`.** It must satisfy the
   statement above exactly.
2. **Do not redefine or shadow anything from the problem's Statement module.**
   Declaring your own `goal`, or redefining a name it depends on, produces a
   proof of a *different* statement and is rejected. This is the single most
   common rejection.
3. **No `sorry`** anywhere, including in helper lemmas. It surfaces as the
   axiom `sorryAx` and is detected transitively through imports.
4. **No `native_decide`** — it surfaces as `Lean.ofReduceBool` and is not
   accepted, because it trusts compiled code rather than the kernel.
5. Only these axioms are permitted: `propext`, `Classical.choice`,
   `Quot.sound`.
6. Follow Mathlib style: hypotheses left of the colon, explicit types,
   `snake_case` theorem names, `UpperCamelCase` types.

## What I want from you

Here is my argument in informal mathematics:

> [PASTE YOUR PROOF SKETCH HERE]

Turn it into Lean 4 that compiles under the environment above and satisfies the
statement exactly. Where you are unsure a lemma exists in this Mathlib version,
say so explicitly rather than guessing a name.

Submissions are not open yet

Conjectura is in beta. You can read every statement, every definition and the Lean behind them, and download the exact files the checker uses — but proofs are not being accepted yet.

The reason is a deliberate order of operations. Accepting a proof means running a stranger’s code and standing behind a verdict, and no statement here yet carries a researcher’s name. A machine-checked answer to a question nobody has vouched for is worth very little, so the vouching comes first.

The English write-ups are also switched off during the beta. Nothing on this page is generated by a model.

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