FormalConjectures/ErdosProblems/1090.lean

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L4DEL explanations are a separate layer. They do not change the Lean source. Intuition is explanatory prose; numbered items are a reading path; sources link the wording back to evidence.

L4DEL
Lean4 Domain Explanation Language

Credit: The upstream authors and contributors retain credit for the project and mathematics. lean4-explainer only renders their pinned source and adds the separately disclosed explanation layer.

Project sources: Pinned project documentation, Formal Conjectures paper, arXiv v1

1 declarations, 1 contain sorry, 0 declared as axioms

Erdős Problem 1090

References:

Imports (1)
  1. theorem Erdos1090.erdos_1090 Let $k\geq 3$ sorry line 38 source

    Source documentation

    Let $k\geq 3$. Does there exist a finite set $A\subset \mathbb{R}^2$ such that, in any $2$-colouring of $A$, there exists a line which contains at least $k$ points from $A$, and all the points of $A$ on the line have the same colour?

    Erdős [Er75f] says Graham and Selfridge proved the answer is yes when $k=3$. Hunter has observed that, for sufficiently large $n$, a generic projection of $[k]^n$ into $\mathbb{R}^2$ has this property, by the Hales-Jewett theorem.

    depends on sorry what this means

    Statement uses: Iff, True, Nat, LE.le, OfNat.ofNat, Exists, Finset, Fin, Real, Subtype, Membership.mem, And, Collinear, GE.ge, Finset.card, AffineSubspace, affineSpan, Eq, Subtype.mk

    Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous

    40 instances, coercions and auxiliary constants left out.

    Tactic trace: 1 steps

    Open to load the checked before-and-after goal states.

    theorem erdos_1090 : answer(True) ↔ ∀ (k : ℕ), ∀ (hk : 3 ≤ k),
        ∃ (A : Finset (Fin 2 → ℝ)), ∀ (C : A → Fin 2),
          ∃ (S : Finset (Fin 2 → ℝ)), ∃ (hSA : S ⊆ A),
            Collinear ℝ (S : Set (Fin 2 → ℝ)) ∧ S.card ≥ k ∧
              (∀ y ∈ A, y ∈ affineSpan ℝ (S : Set (Fin 2 → ℝ)) → y ∈ S) ∧
              ∃ c, ∀ x (hx : x ∈ S), C ⟨x, hSA hx⟩ = c := by
    …

    Mentions: c