Baer–Specker group

DefinitionBergman's question on subgroups of the Baer–Specker group (Kourovka 17.24)#✓fully proved

The proposition that the Baer–Specker group \(\mathbb{Z}^{\mathbb{N}}\) of all integer sequences has a subgroup \(H\) whose dual \(\operatorname{Hom}(H, \mathbb{Z})\) is a free abelian group of rank greater than \(\aleph_0\).

def baerSpeckerSubgroupDualFreeUncountable : Prop :=
  ∃ H : AddSubgroup (ℕ → ℤ), Module.Free ℤ (H →+ ℤ) ∧ Cardinal.aleph0 < Module.rank ℤ (H →+ ℤ)

Used by 1

Used byA subgroup of the Baer–Specker group with free dual of uncountable rank

TheoremA subgroup of the Baer–Specker group with free dual of uncountable rank#●ready to prove

The Baer–Specker group \(\mathbb{Z}^{\mathbb{N}}\) of all integer sequences has a subgroup \(H\) whose dual \(\operatorname{Hom}(H, \mathbb{Z})\) is a free abelian group of rank greater than \(\aleph_0\); this answers Kourovka 17.24 positively.

Bergman's subgroup has dual free abelian of rank \(2^{\aleph_0}\).

public theorem_wanted baerSpeckerSubgroupDualFreeUncountable_holds :
    baerSpeckerSubgroupDualFreeUncountable

Sources

• E. I. Khukhro and V. D. Mazurov (eds.), Unsolved Problems in Group Theory: The Kourovka Notebook, 21st issue, arXiv:1401.0300v49 (2026), Archive of Solved Problems, Problem 17.24 (G. M. Bergman); resolution: G. M. Bergman, More Abelian groups with free duals, Portugaliae Math. 69 (2012), 69-84 (arXiv:1104.3827), abstract (a subgroup of ℤ^ω whose dual Hom(G, ℤ) is free abelian of rank 2^ℵ₀); Kourovka Archive, Problem 17.24 (Yes, it does), https://arxiv.org/abs/1104.3827.

Uses 1

Statement usesBergman's question on subgroups of the Baer–Specker group (Kourovka 17.24)

1 fully proved1 ready to prove

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