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.

  • 1
    $\begingroup$ After a quick glance it looks OK to me. $\endgroup$ – Pedro Sánchez Terraf Feb 4 at 17:55
  • $\begingroup$ Thank you for your time and comment :-) $\endgroup$ – LBJFS Feb 4 at 19:03

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