Let $K$ be the subgroup of $\mathbb{Z}^\mathbb{Z}$ consisting of those functions $f : \mathbb{Z} \to \mathbb{Z}$ with finite image. Is $K$ free abelian?

My guess is no, because $K$ feels too much like the Baer-Specker group $\mathbb{Z}^\mathbb{Z}$, which is not free abelian. However, it's not obvious to me how to adapt the arguments in the proof of Baer-Specker to this case.

If it helps, $K$ also has the following presentation: it is the abelian group generated by the subsets of $\mathbb{Z}$, modulo the relations $\varnothing = 0$ and $A + B = (A \cup B) + (A \cap B)$ for all $A, B \subseteq \mathbb{Z}$. The class of $A \subseteq \mathbb{Z}$ in this presented group corresponds to the indicator function of $A$ in the above definition of $K$.

Another note: $\operatorname{Hom}(K,\mathbb{Z})$ is the group of finitely-additive $\mathbb{Z}$-valued measures on $\mathbb{Z}$ with the discrete $\sigma$-algebra. Maybe this dual group is free abelian of countable rank? In which case it would follow that $K$ is not free.

All thoughts appreciated!


1 Answer 1


Yes this group is free abelian. Actually, for every set $X$ the additive group of bounded functions $X \to \mathbb{Z}$ is free abelian.

Nöbeling, Verallgemeinerung eines Satzes von Herrn E. Specker, Invent. Math. 6 (1968) 41–55.

See also

Andreas Blass, Specker's Theorem for Nöbeling's Group, Proceedings of the American Mathematical Society Vol. 130, No. 6 (Jun., 2002), pp. 1581-1587 (7 pages)

You can find Andreas Blass' paper on his website, and the paper by Nöbeling here.

  • $\begingroup$ Thank you! I am pleasantly surprised :) $\endgroup$ Commented Dec 30, 2023 at 17:10
  • 6
    $\begingroup$ If you are interested, a special case of this is formalized in Lean. $\endgroup$
    – Ricky
    Commented Dec 30, 2023 at 18:16

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