105.lean`

Explanation legend Lean4 Domain Explanation Language (L4DEL)

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L4DEL: Lean4 Domain Explanation Language

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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

4 declarations, 4 contain sorry, 0 declared as axioms

Erdős Problem 105

References:

erdosproblems.com/105
[Be83] Beck, József, On the lattice property of the plane and some problems of Dirac, Motzkin and Erdős in combinatorial geometry. Combinatorica (1983), 281-297.
[ErPu95] Erdős, P. and Purdy, G., Two combinatorial problems in the plane. Discrete Comput. Geom. (1995), 441-443.
[SzTr83] Szemerédi, Endre and Trotter, Jr., William T., Extremal problems in discrete geometry. Combinatorica (1983), 381-392.

Imports (1)

FormalConjecturesUtil

theorem Erdos105.erdos_105 Let $A,B\subset \mathbb{R}^2$ be disjoint sets of size $n$ and $n-3$ respectively, with not all of $A$ contained on a single line sorry line 48 source
Source documentation
Let $A,B\subset \mathbb{R}^2$ be disjoint sets of size $n$ and $n-3$ respectively, with not all of $A$ contained on a single line. Is there a line which contains at least two points from $A$ and no points from $B$?
This has been disproved by Xichuan in the comments, who has found three explicit counterexamples.

depends on sorry what this means
Statement uses: Iff, False, Finset, EuclideanSpace, Real, Fin, OfNat.ofNat, Nat, Disjoint, Eq, Finset.card, HAdd.hAdd, Not, Collinear, ENNReal, Exists, And, Membership.mem, Ne, AffineSubspace and 4 more
Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous
78 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
```
theorem erdos_105 : answer(False) ↔
    ∀ A B : Finset ℝ², Disjoint A B → A.card = B.card + 3 →
      ¬ Collinear ℝ (A : Set ℝ²) →
      ∃ p ∈ A, ∃ q ∈ A, p ≠ q ∧ ∀ b ∈ B, b ∉ line[ℝ, p, q] := by
…
```
theorem Erdos105.erdos_105.variants.hickerson A construction of Hickerson shows that this fails with $n-2$ sorry line 58 source
Source documentation
A construction of Hickerson shows that this fails with $n-2$.

depends on sorry what this means
Statement uses: Not, Finset, EuclideanSpace, Real, Fin, OfNat.ofNat, Nat, Disjoint, Eq, Finset.card, HAdd.hAdd, Collinear, ENNReal, Exists, And, Membership.mem, Ne, AffineSubspace, affineSpan, Insert.insert and 2 more
Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous
78 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
```
theorem erdos_105.variants.hickerson : ¬ ∀ A B : Finset ℝ², Disjoint A B →
    A.card = B.card + 2 → ¬ Collinear ℝ (A : Set ℝ²) →
    ∃ p ∈ A, ∃ q ∈ A, p ≠ q ∧ ∀ b ∈ B, b ∉ line[ℝ, p, q] := by
…
```
theorem Erdos105.erdos_105.variants.beck_szemeredi_trotter A result independently proved by Beck [Be83] and Szemerédi and Trotter [SzTr83] (see [211]) implies it is true with $n-3$ replaced by $cn$ for some constant $c>0$ sorry line 68 source
Source documentation
A result independently proved by Beck [Be83] and Szemerédi and Trotter [SzTr83] (see [211]) implies it is true with $n-3$ replaced by $cn$ for some constant $c>0$.

depends on sorry what this means
Statement uses: Exists, Real, And, GT.gt, OfNat.ofNat, Finset, EuclideanSpace, Fin, Nat, Disjoint, LE.le, Finset.card, HMul.hMul, Not, Collinear, ENNReal, Membership.mem, Ne, AffineSubspace, affineSpan and 3 more
Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous
90 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
```
theorem erdos_105.variants.beck_szemeredi_trotter : ∃ c > (0 : ℝ),
    ∀ A B : Finset ℝ², Disjoint A B → (B.card : ℝ) ≤ c * A.card →
      ¬ Collinear ℝ (A : Set ℝ²) →
      ∃ p ∈ A, ∃ q ∈ A, p ≠ q ∧ ∀ b ∈ B, b ∉ line[ℝ, p, q] := by
…
```
Mentions: c
theorem Erdos105.erdos_105.variants.sub_four It remains possible that this holds with $n-4$ (or in general with $n-O(1)$ or $(1-o(1))n$) sorry line 78 source
Source documentation
It remains possible that this holds with $n-4$ (or in general with $n-O(1)$ or $(1-o(1))n$).

depends on sorry what this means
Statement uses: Iff, True, Finset, EuclideanSpace, Real, Fin, OfNat.ofNat, Nat, Disjoint, Eq, Finset.card, HAdd.hAdd, Not, Collinear, ENNReal, Exists, And, Membership.mem, Ne, AffineSubspace and 4 more
Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous
78 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
```
theorem erdos_105.variants.sub_four : answer(sorry) ↔
    ∀ A B : Finset ℝ², Disjoint A B → A.card = B.card + 4 →
      ¬ Collinear ℝ (A : Set ℝ²) →
      ∃ p ∈ A, ∃ q ∈ A, p ≠ q ∧ ∀ b ∈ B, b ∉ line[ℝ, p, q] := by
…
```