Symmetric hulls of nowhere dense convex sets

DefinitionMazur's symmetric hull claim (Scottish Book Problem 65)#✓fully proved

The claim that for every real Banach space \(E\) and every convex set \(W \subseteq E\) with \(0 \in W\), if the closure of \(W\) has empty interior (that is, \(W\) is nowhere dense), then the closure of the convex hull of \(W \cup (-W)\) also has empty interior. The convex hull of \(W \cup (-W)\) is the smallest convex set containing \(W\) that is symmetric about the origin.

The typescript's parenthetical identifies this set with the difference set \(W - W\), which is larger in general; the smallest-set reading is adopted here. For the Ad 65 counterexample \(W\) (non-decreasing functions, a cone) the two constructions coincide, so the counterexample still applies.

def ScottishBookProblem65Claim : Prop :=
  ∀ (E : Type) [NormedAddCommGroup E] [NormedSpace ℝ E] [CompleteSpace E]
    (W : Set E),
    Convex ℝ W → 0 ∈ W → interior (closure W) = ∅ →
    interior (closure (convexHull ℝ
      (W ∪ {z | ∃ w ∈ W, z = -w}))) = ∅
Used by 1

TheoremMazur's Problem 65 has a negative answer#●ready to prove

The symmetric hull claim of Scottish Book Problem 65 is false: there are a real Banach space \(E\) and a convex set \(W \subseteq E\) with \(0 \in W\) whose closure has empty interior, such that the closure of the convex hull of \(W \cup (-W)\) has nonempty interior.

theorem_wanted mazur65_refuted : ¬ ScottishBookProblem65Claim
Sources
  • S. Mazur, Problem 65 of the Scottish Book; typescript Ad 65 records the negative answer, with the counterexample of the set W of non-decreasing functions in the space (C) of continuous functions: W is convex, nowhere dense, and contains 0, while the symmetric convex set containing W contains all functions of bounded variation and is not nowhere dense (indeed, it is dense, since it contains the polynomials). For this W (a cone) the W ∪ -W hull and the difference-set construction coincide, so the counterexample refutes the corrected claim too.
Uses 1
1 fully proved1 ready to prove

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