# Is this group free abelian?

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!

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