FormalConjectures/GreensOpenProblems/3.lean
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1 declarations, 1 contain sorry, 0 declared as axioms
Ben Green's Open Problem 3
References:
- [Ben Green's Open Problem 3](https://people.maths.ox.ac.uk/greenbj/papers/open-problems.pdf#section.3 Problem 3)
- [FGY26] Franchi, Leonardo and Gowers, W. T. and Yip, Fredy, Product-free subsets of $(0,1)$, arXiv:2607.06073, Theorem 1.1.
Imports (1)
- theorem
Green3.green_3Suppose that $A \subset [0,1]$ is open and has measure greater than $\frac{1}{3}$ sorry line 41 sourceSource documentation
Suppose that $A \subset [0,1]$ is open and has measure greater than $\frac{1}{3}$. Is there a solution to $xy = z$ with $x, y, z \in A$?
The answer is yes. [FGY26] prove that an open product-free subset of $(0,1)$ has measure strictly less than $\frac{1}{3}$.
depends on sorry what this means
Statement uses:
Iff,True,Set,Real,IsOpen,LE.le,Set.Icc,OfNat.ofNat,GT.gt,ENNReal,MeasureTheory.Measure,MeasureTheory.MeasureSpace.volume,HDiv.hDiv,Nat,Exists,And,Membership.mem,Eq,HMul.hMulProof uses:
sorryAx,Lean.Name,Bool.false,Lean.Name.num,Lean.Name.str,Lean.Name.anonymous60 instances, coercions and auxiliary constants left out.
Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem green_3 : answer(True) ↔ ∀ A : Set ℝ, IsOpen A → A ⊆ Icc 0 1 → volume A > 1/3 → ∃ x y z, x ∈ A ∧ y ∈ A ∧ z ∈ A ∧ x * y = z := by …