Elements of $\operatorname{Hom}_{\mathbb{Z}} (\mathbb{Q}, \mathbb{Q}/\mathbb{Z}) $ as Cauchy sequences

Question

There is an isomorphism of abelian groups $\operatorname{Hom}_{\mathbb{Z}} (\mathbb{Q}, \mathbb{Q}/\mathbb{Z}) \cong \mathbb{R}$, a proof is based on three observations:

1. $\operatorname{Hom}_{\mathbb{Z}} (\mathbb{Q}, \mathbb{Q}/\mathbb{Z}) $ is torsion free.

2. $\operatorname{Hom}_{\mathbb{Z}} (\mathbb{Q}, \mathbb{Q}/\mathbb{Z}) $ is a divisible abelian group.

3. Hence $\operatorname{Hom}_{\mathbb{Z}} (\mathbb{Q}, \mathbb{Q}/\mathbb{Z}) $ is a vector space over $\mathbb{Q}$ and it has the same cardinality as $\mathbb{R}$, hence there is an isomorphism of $\mathbb{Q}$-vector spaces $\operatorname{Hom}_{\mathbb{Z}} (\mathbb{Q}, \mathbb{Q}/\mathbb{Z}) \cong \mathbb{R}$.

This proof is indirect. If we think about elements of $\mathbb{R}$ as classes of Cauchy sequences of elements of $\mathbb{Q}$, can we directly interpret such sequence as an element in $\operatorname{Hom}_{\mathbb{Z}} (\mathbb{Q}, \mathbb{Q}/\mathbb{Z})$?

Answer 1

$$End(\Bbb{Q/Z})=\varprojlim_{n\to \infty} End(\Bbb{n^{-1}Z/Z})= \varprojlim_{n\to \infty} \Bbb{Z/nZ}$$ (the profinite integers)

$$Hom_{\mathbb{Z}} (\mathbb{Q}, \mathbb{Q}/\mathbb{Z})=\Bbb{Q\otimes_Z}End(\Bbb{Q/Z})=\Bbb{A_{Q,fin}}$$ (the finite adeles)

It is well-known that the $\Bbb{Q}$-vector space isomorphisms or embeddings $\Bbb{A_{Q,fin}}\to \Bbb{R}$ need the axiom of choice and that none has a concrete expression (no idea in what theory we can prove it, if we can).

$\Bbb{A_{Q,fin}}$ is the completion of $\Bbb{Q}$ for the norm $$\|\pm \prod_j p_j^{e_j}\|=\sup_j p_j^{-e_j}, \qquad e_j\in \Bbb{Z}$$