FormalConjectures/ErdosProblems/1090.lean
Explanation legend Lean4 Domain Explanation Language (L4DEL)
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:
- erdosproblems.com/1090
- [Er75f] Erdős, Paul, On some problems of elementary and combinatorial geometry. Ann. Mat. Pura Appl. (4) (1975), 99-108.
Imports (1)
- theorem
Erdos1090.erdos_1090Let $k\geq 3$ sorry line 38 sourceSource 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.mkProof uses:
sorryAx,Lean.Name,Bool.false,Lean.Name.num,Lean.Name.str,Lean.Name.anonymous40 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