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LeanPool.ConwayRefinement.ConwayRefinement.Topology.Order.Tests.CantorBendixsonConvexCover

The Cantor–Bendixson convex cover has a nondegenerate model #

The partition hypotheses require a nested well-ordered neighborhood base at zero consisting of open convex subgroups. The real line does not satisfy them: its only convex subgroups are zero and the whole line, and zero is not open. Any witness is therefore non-Archimedean, and this file supplies one: real sequences indexed by the naturals under the lexicographic order, with the subgroups of sequences vanishing below a given index.

Those subgroups are nested, convex, open, and coinitial, so the hypotheses are consistent and the geometric part of the cofactor construction is not vacuous. The check also separates the openness requirement from the trivial family: the zero subgroup alone satisfies every other condition.

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Real sequences under the lexicographic order: a non-Archimedean ordered abelian group.

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    The entries of a lexicographic sequence.

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      theorem Tests.CantorBendixsonConvexCover.lt_of_forall_eq_of_lt {x y : LexSeq} {m : ℕ} (h : ∀ j < m, entry x j = entry y j) (hm : entry x m < entry y m) :
      x < y

      The subgroup of sequences vanishing below a given index.

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        The witnessing family is nested and decreasing.

        theorem Tests.CantorBendixsonConvexCover.exists_least_ne_zero {c : LexSeq} {k : ℕ} (hne : entry c k ≠ 0) :
        ∃ (m : ℕ), entry c m ≠ 0 ∧ ∀ j < m, entry c j = 0

        The least index at which a nonzero sequence does not vanish.

        Each subgroup of the family is convex: a sequence between two vanishing ones has no earlier nonzero entry, since a positive one would exceed the upper bound and a negative one would fall below the lower bound.

        The indicator sequence with a single unit entry.

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          Every sequence vanishing below an index lies strictly between the negative and positive unit sequences at that index, so each subgroup of the family contains a neighborhood of zero and is therefore open.

          The family is coinitial: every strictly positive sequence dominates one of its members.

          theorem Tests.CantorBendixsonConvexCover.exists_disjoint_convex_cover_with_rank_lt_center_lexSeq (s : TopologicalSpace.Closeds LexSeq) (hs : (↑s).IsPWO) :
          ∃ (X : Set LexSeq) (C : ↑X → Set LexSeq), X ⊆ ↑s ∧ (∀ (x : ↑X), ↑x ∈ C x) ∧ (∀ (x : ↑X), IsOpen (C x)) ∧ (∀ (x : ↑X), (C x).OrdConnected) ∧ (∀ (x y : ↑X), x ≠ y → Disjoint (C x) (C y)) ∧ (∀ (x y : ↑X), ↑x < ↑y → ∀ a ∈ C x, ∀ b ∈ C y, a < b) ∧ ↑s ⊆ ⋃ (x : ↑X), C x ∧ (∀ (x : ↑X), ∀ z ∈ ↑s ∩ C x, z ≤ ↑x) ∧ (∀ (x : ↑X), ∀ z ∈ ↑s ∩ C x, z ≠ ↑x → s.cantorBendixsonRank hs z < s.cantorBendixsonRank hs ↑x) ∧ ∀ (z : LexSeq), ¬AccPt z (Filter.principal X)

          The cover hypotheses are consistent. Every hypothesis of the disjoint convex cover theorem holds for the lexicographic sequence group with the vanishing-below family, so the result is not vacuous.

          The real line is not a witness: its only convex subgroups are zero and the whole line, and zero is not open, so no family of open convex subgroups is a neighborhood base at zero. The openness requirement is therefore doing real work.