Work in ZF (so no choice). Then it is consistent that there is no (Hamel) basis for $\mathbb{R}$ as a $\mathbb{Q}$-vector space. My question is about models where $\mathbb{R}$ does have a basis, but choice still fails in terrible ways. Specifically:

Is it consistent with ZF that there is a basis for $\mathbb{R}$ as a $\mathbb{Q}$-vector space which is Borel?

(This question arose out of Is there any uncountably infinite set that does not generate the reals?, my answer, and Ricky Demer's and my comments to it.)

Some observations:

  • By "Borel" I mean the weak sense of Borel: "in the smallest $\sigma$-algebra containing the open sets." Note that if we use the stronger (and better-behaved) notion "Borel coded", meaning "there is a Borel code for", then the answer to the question is easily no.

  • How bad can Borel-ness be without choice? Well, it is consistent with ZF that every set of reals is Borel; specifically, it's consistent with ZF that $\mathbb{R}$ is a countable union of countable sets.

  • However, such models don't seem to answer the question; as far as I know, in such models, $\mathbb{R}$ doesn't have a basis.

  • So the question becomes: can we on the one hand "stretch" the notion of Borel-ness by killing choice very badly, while preserving the existence of a basis in the first place (so forcing us to not kill choice too badly)?

NOTE: This question is related in spirit (if not content) to my previous question Spanning the reals with a small set - choicelessly. Specifically, I mention this because I think techniques useful to one may be useful to the other.

  • $\begingroup$ Well. "Every Borel set has a code" is equivalent (if I recall correctly) to countable choice on sets of reals. So at least under $\sf AC_\omega(\Bbb R)$ (or something like that), we can easily conclude that the answer has to be negative as well. So something must be seriously broken in the real numbers for this to happen. $\endgroup$
    – Asaf Karagila
    Commented Mar 16, 2016 at 19:47
  • 1
    $\begingroup$ @AsafKaragila "Something must be seriously broken" is what I like to hear! :D $\endgroup$ Commented Mar 16, 2016 at 20:05

2 Answers 2


This is a hard question. If only for the fact that we don't really know any models of $\sf ZF$ in which $\Bbb R$ is not well-ordered and there is a Hamel basis for $\Bbb R$ over $\Bbb Q$.

While in this person's opinion, the existence of a Hamel basis should not be equivalent to the fact that $\Bbb R$ is well-ordered, but I am not sure how to prove something like that.

  • $\begingroup$ Re: your last sentence, indeed that question seems to be open: math.stackexchange.com/questions/122857/…. $\endgroup$ Commented Mar 16, 2016 at 20:09
  • $\begingroup$ This appears to be more of a comment, although I do appreciate your input. $\endgroup$
    – user223391
    Commented Mar 16, 2016 at 20:09
  • $\begingroup$ Actually, out of curiosity: is it obvious that in the models where $\mathbb{R}$ is countable-by-countable, it does not have a Hamel basis? Clearly it shouldn't, but I'm not familiar enough with the construction to see it. $\endgroup$ Commented Mar 16, 2016 at 20:10
  • $\begingroup$ @ZacharySelk I suspect the reason this is an answer as opposed to a comment is that Asaf is suggesting that the problem is probably open. $\endgroup$ Commented Mar 16, 2016 at 20:11
  • $\begingroup$ (Asaf I just realized the answer I cited a couple comments ago was yours. :P) $\endgroup$ Commented Mar 16, 2016 at 20:15

Well, if you assume a little bit of choice($\mathsf{DC}$), the answer is no, because of the following argument.

So suppose $B$ is a Hamel basis of $\Bbb R$ over $\Bbb Q$ that also happens to be a Borel set. For each $n\geq 1$, and each $\bar q=(q_1,\ldots,q_n)\in\Bbb Q^n,$ consider the set

$$B_{\bar q}:=q_1B+\cdots+q_nB,$$

then $B_{\bar q}$ is an analytic set, as it is the direct image of the Borel set $B^n$ via the continuous function $\Bbb R^n\rightarrow \Bbb R$ given by $(x_1,\ldots,x_n)\mapsto q_1x_1+\cdots+ q_nx_n$. We have $$\Bbb R=\bigcup_{\bar q\in\Bbb Q^{<\omega}}B_{\bar q}.$$

Thus as this is a countable union and analytical sets are Lebesgue measurable, we must have that $\mu(B_{\bar q})>0$ for some $\bar q\in\Bbb Q^{<\omega}$. Because of this result we know $$B_{\bar q\frown\bar q}=B_{\bar q}+B_{\bar q}$$ contains an open interval. We may assume this open interval contains $0$ using a function $\Bbb R\rightarrow\Bbb R$ of the form $x\mapsto x+a$ , so that $$\bigcup_{n\geq 1}nB_{\bar q\frown\bar q}=\Bbb R,$$

which implies that any $x\in\Bbb R$ is of the form $$m_1q_1x_1+\cdots+m_nq_nx_n+m_1'q_1y_1+\cdots+m_n'q_ny_n,$$

for some $x_1,\ldots,x_n,y_1,\ldots,y_n\in B$ and some $m_1,\ldots,m_n,m_1',\ldots,m_n'\in\Bbb N$ and this is a contradiction, as $\Bbb Q$ is not a finitely generated abelian group. Therefore there cannot exist such set $B.$

Under the presence of large cardinals and using the axiom of choice there cannot even exist Hamel basis that are projective.

If there is for instance a supercompact cardinal, there exists an elementary embedding $$j:L(\Bbb R)\rightarrow L(\Bbb R)^{Col(\omega,<\kappa)},$$ whenever $\kappa$ is a supercompact cardinal; this is a theorem you can find in Woodin's article Supercompact cardinals, sets of reals, and weakly homogeneous trees. Such embeddings can be gotten using way weaker assumptions.

Here's why:

We can take $L(\Bbb R)^{Col(\omega,<\kappa)}$ as the Solovay's model as $\kappa$ is inaccesible. Because of this MO answer there is no Hamel basis of $\Bbb R$ as a $\Bbb Q$-vector space in $L(\Bbb R)^{Col(\omega,<\kappa)}$. Hence for any $X\in L(\Bbb R)$ we have that in $L(\Bbb R)$, $X$ is not a Hamel basis of $\Bbb R$ over $\Bbb Q$, using the elementarity of $j$.

But $\Bbb R\subset L(\Bbb R)$ and $\Bbb R\in L(\Bbb R)$, hence $X$ is a Hamel basis of $\Bbb R$ over $\Bbb Q$ in $L(\Bbb R)$ if and only it is in $V$. However all projective sets belong to $L(\Bbb R)$, thus no Hamel basis can be projective.

  • $\begingroup$ Lovely! Although, how exactly are we defining a supercompact in the absence of choice? It's not clear to me that there isn't a problem here . . . Also, does Woodin's result hold even in a $V$ in which choice fails badly? $\endgroup$ Commented Mar 18, 2016 at 6:20
  • $\begingroup$ Well, in Solovay's model you have DC, so every Borel set has a code. $\endgroup$
    – Asaf Karagila
    Commented Mar 18, 2016 at 13:00
  • $\begingroup$ @AsafKaragila "in Solovay's model you have DC" In general, without choice in $V$? How are we defining inaccessibles here? (Although DC holds in $L(\mathbb R)$ under $L(\mathbb R)$-determinacy.) $\endgroup$ Commented Mar 18, 2016 at 15:06
  • $\begingroup$ @Andrés: Well, good question, but I think the answer is positive. $\endgroup$
    – Asaf Karagila
    Commented Mar 18, 2016 at 15:17
  • $\begingroup$ @NoahSchweber, please check my edit. $\endgroup$ Commented Jun 9, 2016 at 22:21

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .