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?
Thank you in advance for your help.