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.

  • $\begingroup$ 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. $\endgroup$ – Nate Eldredge Sep 27 '19 at 3:01
  • $\begingroup$ @Nate If I'm not wrong, there is no (explicit) construction here ams.org/journals/jams/2003-16-01/S0894-0347-02-00409-5/… (see p.15/26, row -14). Maybe I miss something obvious, otherwise I don't understand how get it is Borel. $\endgroup$ – LBJFS Sep 27 '19 at 21:05
  • $\begingroup$ @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.) $\endgroup$ – LBJFS Sep 27 '19 at 21:09
  • $\begingroup$ 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. $\endgroup$ – Nate Eldredge Sep 28 '19 at 1:41

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