I'm wondering whether $(\mathbb{D}, +)$ the additive group of the dyadic rationals is isomorphic to a free Abelian group or not.
On an intuitive level, it seems like it's obviously nonfree, but I'm wondering how you prove it.
For any given finite set $X \subset \mathbb{D}$, $X$ does not generate $\mathbb{D}$. We can prove this by picking $\frac{1}{2^{1+n}}$ where $2^n$ is the largest denominator in $X$.
We know that $|X| \le \aleph_0$ by a cardinality argument.
This leaves countably infinite $X$.
If $X \subset \frac{1}{2^\mathbb{N}}$ and $X$ generates $\mathbb{D}$, then, since $X$ is not finite, any element of $X$ can be generated by adding together the next smaller element an appropriate number of times.
So far, so good.
The step that I'm struggling with is the other countably infinite cases.
If $X$ is countably infinite and has an upper bound on the size of its denominators, then it can't generate $\mathbb{D}$.
Suppose $X$ is countably infinite and has no upper bound on the size of its denominators.
I'm not sure how to show that any fixed $x \in X$ is a finite $\mathbb{Z}$-linear combination of elements taken from $\{ y : y \in X \land y < x \}$.