A point in the Euclidean plane ℝ², modelled as EuclideanSpace ℝ (Fin 2).
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IsDeltaTube T δ says that T is a δ-tube, i.e. it is of the form
deltaTube c v δ for some centre c and unit-length direction v.
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A finite set E ⊂ B(0,1) of N points is "well-spaced" (with constant
C) in the sense of Sharp Projection Theorems III: it lies in the unit
ball, has cardinality comparable to N, and every ball of radius N^{-1/2}
contains at most C points of E.
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Geometric incidence fact: a pair of points x₁, x₂ separated by at
least δ can lie in at most one δ-tube. (Two distinct δ-tubes through
the same δ-separated pair would force their directions to coincide.)
A family of δ-tubes 𝕋 contained in a fixed ρ-tube Tρ, each of
which contains many points of E and additionally satisfies the
two-ends condition: for each T ∈ 𝕋 one can split its ≳ R points
of E ∩ T into two clusters H₁ T, H₂ T, both of cardinality at least
C⁻¹ R, separated by Euclidean distance at least δ. This is exactly
the hypothesis of the two-ends lemma.
- hTρ_tube : IsDeltaTube Tρ ρ
- is_tube (T : Set Point) : T ∈ self.tubes → IsDeltaTube T δ
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Combinatorial consequence of pair_determines_tube_geometric: a fixed
ordered pair (x₁, x₂) can play the role of (H₁, H₂) representatives for
at most one tube in the family. This is the pointwise upper bound on the
double-counting sum used in the proof of the two-ends lemma.
Lemma (two ends). Let E be a well-spaced set in B¹ ⊂ ℝ² with
|E| ∼ N and |E ∩ B(x, N^{-1/2})| ≲ 1, let δ ≤ ρ ≤ N^{-1/2}, and let
Tρ be a ρ-tube with |Tρ ∩ E|_ρ ∼ R̃. If every δ-tube
T ∈ 𝕋_R(E, Tρ) obeys the two-ends condition, then
$$\big|\mathbb{T}_R(E, T_\rho)\big| \;\lesssim\; \frac{\tilde R^{\,2}}{R^{\,2}}.$$
The proof double-counts triples (T, x, y) with x ∈ H₁ T, y ∈ H₂ T,
using pair_determines_tube_from_geometry for the upper bound.