What does it mean to choose a sequence in a Borel fashion?

I have to face with the following descriptive set-theoretic fact, which I don't understand.

Here are some preliminaries. Define a torsion-free abelian group $$A$$ of rank $$n\ge 1$$ to be $$p$$-local for some prime (integer) $$p$$ iff $$A=qA$$ for every other prime $$q$$, or equivalently, iff $$A$$ is a $$\Bbb{Z}_{(p)}$$-module, where $$\Bbb{Z}_{(p)}$$ is the set of all rationals whose denominator is prime to $$p$$.

Now, consider the space $$R^{(p)}(\Bbb{Q}^n)$$ of all $$p$$-local torsion-free abelian groups of rank $$n\ge1$$ and say that a sequence $$(a_1,\dots,a_l)$$ of nonzero elements of $$A$$ is $$p$$-independent iff whenever $$n_1,\dots,n_l\in\Bbb{Z}$$ are s.t. $$n_1a_1+\dots+n_la_l\in pA$$, then $$p$$ divides $$n_j$$ for all $$j=1,\dots,l$$. The sequence $$(a_1,\dots,a_l)$$ is a $$p$$-basis iff it is a maximal $$p$$-independent sequence.

We already know that $$R^{(p)}(\Bbb{Q}^n)$$ is a Borel subset of the power set $$\mathcal{P}(\Bbb{Q}^n)$$ equipped with the product topology via the natural bijection with $$2^{\Bbb{Q}^n}$$; in other words, it is a standard Borel space.

Here is my question:

Fix some $$A\in R^{(p)}(\Bbb{Q}^n)$$. At this point, it is claimed that "we can clearly choose a $$p$$-basis $$(a_1,\dots,a_l)$$ of $$A$$ in a Borel fashion" and I don't understand what does it mean.

As far as I understand, we cannot even assign such a sequence in a unique way, for fixed $$A$$, so we are not talking about a "function" (am I wrong?). Further, why could I choose it in a Borel way?

• If each admissible group is identified with a subset of $\mathbb{Q}^n$, then a basis of such a group would be identified with a subset of that subset, which is again a subset of $\mathbb{Q}^n$. So I would interpret this statement as "there exists a Borel map $F : R^p(\mathbb{Q}^n) \subset 2^{\mathbb{Q}^n} \to 2^{\mathbb{Q}^n}$ such that for each $G \in R^p(\mathbb{Q}^n)$, $F(G)$ is a $p$-basis of $G$ under these identifications". How we prove that, I don't know enough to say, but if there is a standard way of constructing a $p$-basis, I would look at that construction. – Nate Eldredge Sep 27 '19 at 3:01
• @Nate About the construction of $F\colon R^p(\Bbb{Q}^n)\to 2^{\Bbb{Q}^n}$, is this a function? I'm not sure it is. My idea is that if, for example, we consider the space of $\Bbb{Q}$-subspaces of $\Bbb{Q}^n$ of dimension $1\le k< n$, we have that there is no unique way to assign a basis to a subspace (there are actually infinitely many.) – LBJFS Sep 27 '19 at 21:09
• Yes, I mean $F$ here to be a function - it associates each group to some basis for that group. When such a basis is not unique, then the function needs to "choose" one. Again, this is not my area of expertise, so I can't comment on how that might actually be done. But one note is that you could fix a bijection that "labels" $\mathbb{Q}^n$ by $\mathbb{N}$, and whenever you need to choose an "arbitrary" group element meeting some conditions, you choose the one with the lowest-numbered label. – Nate Eldredge Sep 28 '19 at 1:41