A vertex of a polytope stays a vertex of any relaxation keeping its tight rows
ProvedHirsch.relaxation_vertexLet , let be a set of row indices, and let be any set on which the inequalities indexed by hold:
Let be a vertex (extreme point) of with , and suppose every inequality tight at has its index in . Then is an extreme point of .
This is the "vertices survive the dropping of inequalities they do not touch" step of Kalai--Kleitman, stated for an arbitrary superset (so that it applies to a relaxation intersected with a supporting hyperplane and with an auxiliary bounding cut). The reason is that the tight rows at contain linearly independent normals; on an open segment through inside those rows are at the endpoints and at , hence tight at both endpoints, which forces the endpoints to coincide with .
Formalization Note The hypothesis on is only that the -rows hold on ; need not be a polyhedron, and is not assumed (only ).
import Mathlib import Definitions.Def_Hirsch_model open scoped RealInnerProductSpace
namespace Hirsch
theorem relaxation_vertex (d n : ℕ)
(a : Fin n → EuclideanSpace ℝ (Fin d)) (b : Fin n → ℝ)
(T : Finset (Fin n)) (Q : Set (EuclideanSpace ℝ (Fin d)))
(hQ : ∀ z ∈ Q, ∀ j ∈ T, ⟪a j, z⟫ ≤ b j)
(x : EuclideanSpace ℝ (Fin d)) (hx : x ∈ Set.extremePoints ℝ (Hpoly a b)) (hxQ : x ∈ Q)
(hxT : ∀ j, ⟪a j, x⟫ = b j → j ∈ T) :
x ∈ Set.extremePoints ℝ Q := by sorry
end Hirsch