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17.24 (2010)

Solved

(A. Blass, J. Irwin, G. Schlitt). Does $\mathbb{Z}^\omega$ have a subgroup whose dual is free of uncountable rank?

Contributor: G. M. Bergman

Progress

Yes, it does (G. M. Bergman, Portugaliae Math., 69 (2012) 69–84).

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