Which sets of positive integers are forced to contain long arithmetic progressions? Van der Waerden (1927) showed that in any finite colouring of some colour class does; Erdős and Turán (1936) asked for a density version, which became Szemerédi's theorem. Erdős then proposed the strongest natural size condition: divergence of the reciprocal sum. Erdős Problem #3 (erdosproblems.com/3) asks whether every with contains arbitrarily long arithmetic progressions. Erdős attached one of his largest prizes to it. The primes are the motivating example: , so a positive answer would contain the Green–Tao theorem.
Timeline.
The problem is open for every .
A set is an arithmetic progression of length if and for some (for the size condition forces ). For , denotes the largest size of a subset of containing no arithmetic progression of length . A set has divergent reciprocal sum if .
For every ,
This is the formal-conjectures statement erdos_3 with its answer(sorry) instantiated to the conjectured answer yes. A disproof of the goal on the platform settles the problem negatively.
A positive answer would be a common strengthening of Szemerédi's theorem and the Green–Tao theorem, obtained from a single size condition with no arithmetic structure. Through the reduction milestone, it is closely tied to the quantitative theory of : bounds of the shape for every would suffice. Formalizing the milestones would also give reusable Lean statements of Szemerédi-type theorems in a common language.
Divergence of is a very weak condition: such sets can have density zero, and the natural approach through requires bounds just past . For this barrier was only broken in 2020. For the best known bounds (Leng–Sah–Sawhney) save only a power of in the exponent, far from what is needed. The Green–Tao method uses pseudorandom majorants specific to the primes and does not apply to arbitrary sets.
All statements import the published definition file Erdos142Basic, which reproduces the formal-conjectures definitions IsAPOfLengthWith, IsAPOfLength and the counting function r k N (over ). The reciprocal-sum hypothesis is ¬ Summable (fun a : A ↦ 1 / (a : ℝ)); the element , if present, contributes . "Arbitrarily long" is written as ∃ᶠ k in atTop, which is equivalent to "every length" because sub-progressions of progressions are progressions. Bounds stated in the literature with are written without a multiplicative constant and with "for all sufficiently large "; the constant can be absorbed into the exponent. Contributions formalizing partial summation over sets of naturals and the equivalence of the "frequently" and "for every " forms are welcome.
namespace Erdos3
open Erdos142
theorem erdos_3 : ∀ A : Set ℕ,
(¬ Summable fun a : A ↦ 1 / (a : ℝ)) →
∃ᶠ (k : ℕ) in Filter.atTop, ∃ S ⊆ A, IsAPOfLength S k := by
sorry
end Erdos3Let satisfy . Then contains arbitrarily long arithmetic progressions: for infinitely many (equivalently, for every ) there is a -term progression with all of whose terms lie in .
This is the conjecture of Erdős (Erdős Problem #3), still open. The formal-conjectures entry states it as answer(sorry) ↔ …; here the answer is fixed to Erdős's conjectured answer yes, so a proof on the platform settles it affirmatively and a disproof (a proof of the negation) settles it negatively.
No open leaves. Every sub-goal is proved or awaiting decomposition.