Minimal construction interface for the Wallace counterexample #
The construction supplies characters compatible with all prescribed ultrafilter limits and
separating the points of the free group. For the Wallace corollary it is enough that, for every
nonzero element x, one compatible character does not annihilate x. This file isolates
precisely that output.
No topology on the free Abelian group and no compactness conclusion is stored in
SeparationPackage; both are derived below from its algebraic and filter-theoretic fields.
An injective sequence bundled with the proof of injectivity.
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- Wallace.InjectiveSequence' G = { s : ℕ → G // Function.Injective s }
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The minimal post-fusion data needed for Wallace's problem.
The difficult construction must supply the coding, the prepared subsequences and free ultrafilters, and a compatible separating character for every nonzero element. The structure does not assume any topology or compactness property of the free Abelian group.
- Code : Type u
Codes for injective sequences in the free Abelian group.
The identification of codes with injective sequences.
- codeIndex : self.Code → I
The distinguished coordinate assigned to each code.
The selected subsequence of the sequence represented by each code.
- subsequence_strictMono (c : self.Code) : StrictMono (self.subsequence c)
- ultrafilter : self.Code → Ultrafilter ℕ
The free ultrafilter along which the selected subsequence converges.
A circle-valued character separating each nonzero group element.
- character_limit (c : self.Code) (x : { x : I →₀ ℤ // x ≠ 0 }) : Filter.Tendsto (fun (n : ℕ) => (self.character x) (↑(self.codeEquiv c) (self.subsequence c n))) (↑(self.ultrafilter c)) (nhds ((self.character x) (Finsupp.single (self.codeIndex c) 1)))
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Simultaneous evaluation by all compatible separating characters.
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The initial topology induced by the compatible separating characters.
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The prepared subsequence converges to its prescribed basis point in the initial topology.
The package supplies the exact free-ultrafilter limit property used by the cone argument.