Is there a $\mathbb{Q}$-basis of $\mathbb{R}$ that is Borel?

Question

A question that came to my mind yesterday:

Applying zorn's lemma (and hence, the axiom of choice), one can show that there must be a Hamel basis for the $\mathbb{Q}$-vector space $\mathbb{R}$, i.e. a subset $H\subset \mathbb{R}$, such that any real number $x$ can be uniquely expressed as a finite $\mathbb{Q}$-linear combination of elements of $H$. There is no way to construct such a set $H$, but nevertheless we are still able to describe some properties of $H$, such as that $H$ must be uncountable or that $H$ must contain at most one rational (two rationals wouldn't be linearily independent anylonger) or that $H$ must be totally disconnected (suppose not, then $H$ contains a proper interval and therefore more than one rational). So its seems that everey such $\mathbb{Q}$-bases of $\mathbb{R}$ must have a very odd structure.

Now my question is: of all the possible $\mathbb{Q}$-bases of $\mathbb{R}$, is there at least one that is tame enough to be a borel set? I mean there are after all strange structured borel sets out there and there are at least uncountably many $\mathbb{Q}$-bases of $\mathbb{R}$, but on the other hand, a $\mathbb{Q}$-basis of $\mathbb{R}$ seems to behave too strange to be a borel set..

So what do you think?

Answer 1

It is not possible for a Hamel basis of $\mathbb{R}$ as a $\mathbb{Q}$ vector space to be Borel. See Joel Hamkins' answer here: https://mathoverflow.net/questions/46063/explicit-hamel-basis-of-real-numbers, which also describes some consistency results about how close to Borel such a basis can be.

Answer 2

As a complement to Chris Eagle's answer, we can add at least a bit of regularity can be asked of a Hamel basis of $\mathbb{R}$ over $\mathbb{Q}$: By using Ramsey methods, it can be shown that there is a Cantor set of reals that are linearly independent over the rationals (See Kechris' Classical Descriptive Set Theory, Exercise 19.2). So, there are Hamel bases that include a perfect set.