# Direct sum of subgroups of $\mathbb{Q}^n$ with $\mathbb{Q}$ is a Borel map - Self study

Let $$\mathcal{Pow}(\mathbb{Q}^n)$$ be the power set of $$\mathbb{Q}^n$$ and consider the product topology induced by the natural bijection $$\mathcal{Pow}(\mathbb{Q}^n)\cong 2^{\mathbb{Q}^n}$$ defined by $$A\mapsto \chi_A$$. Hence the induced ("product") topology on $$\mathcal{Pow}(\mathbb{Q}^n)$$ is generated by the basic open sets of the form $$\mathcal{O}_{F,\,G}=\{S\in\mathcal{Pow}(\mathbb{Q}^n)\mid F\subseteq S,\,S\cap G=\emptyset\}$$, where $$F,\,G\subseteq \mathbb{Q}^n$$ are finite. Assume $$R(\mathbb{Q}^n)$$ denotes the set of all subgroups of $$\mathbb{Q}^n$$ of rank $$n$$. It is clearly equipped with the relative topology.

I want to prove the following:

"Let $$f\colon R(\mathbb{Q}^n)\to R(\mathbb{Q}^{n+1})$$ be the map defined by $$f(A)=A\oplus \mathbb{Q}$$. Show that it is Borel."

Here is my attempt: let $$\mathcal{O}_{F,\,G}\cap R(\mathbb{Q}^{n+1})$$ be a (nonempty) basic open subset of $$R(\mathbb{Q}^{n+1})$$. I want to show that $$f^{-1}(\mathcal{O}_{F,\,G}\cap R(\mathbb{Q}^{n+1}))=\{A\in R(\mathbb{Q}^{n})\mid F\subseteq (A\oplus\mathbb{Q}),\, (A\oplus\mathbb{Q})\cap G=\emptyset\}$$ is Borel. Now, if $$\pi\colon\mathbb{Q}^{n+1}\to\mathbb{Q}^n$$ is the projection on the first $$n$$ components, then the following should hold: $$F\subseteq (A\oplus\mathbb{Q})$$ iff $$\pi(F)\subseteq A$$. In a similar way $$(A\oplus\mathbb{Q})\cap G=\emptyset$$ iff $$A\cap \pi(G)=\emptyset$$.

Then $$f^{-1}(\mathcal{O}_{F,\,G}\cap R(\mathbb{Q}^{n+1}))=\{A\in R(\mathbb{Q}^n)\mid \pi(F)\subseteq A,\,A\cap \pi(G)=\emptyset\}$$ is open. In other words, if this is the case, $$f$$ is also continuous.

Is my attempt correct? I think I'm missing something because otherwise continuity should have been stated explicitly.

Thank you in advance for your help.

• After a quick glance it looks OK to me. – Pedro Sánchez Terraf Feb 4 at 17:55
• Thank you for your time and comment :-) – LBJFS Feb 4 at 19:03