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Without (at least a weak form of) choice, we cannot prove the existence of a basis for $\mathbf{R}$ over $\mathbf{Q}$. But is there some vector space $V\supset \mathbf{R}$ that we can prove has a basis, without choice?

Note that it is not true in general that if $V$ has a basis and $U\subseteq V$ then $U$ has a basis; see https://mathoverflow.net/questions/80765/if-v-is-a-vector-space-with-a-basis-w-subseteq-v-has-to-have-a-basis-too

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    $\begingroup$ @Asaf Karagila I think this is not a duplicate of math.stackexchange.com/questions/257325/… because that question gets us "There are models where R has no basis over Q" but that does not get us "There are models where no superspace of R has no basis over Q"; in particular, I do not see why a superspace of R having a basis implies R has a basis. $\endgroup$ – alphacapture Sep 30 '17 at 23:56
  • $\begingroup$ You are absolutely right. I misread the question. $\endgroup$ – Asaf Karagila Oct 1 '17 at 5:58
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No. If there existed such a basis, we could define the following linear function $f:V\to\mathbf{Q}:$

Write $1\in\mathbf{R}$ as a linear combination of the basis vectors. Let $v$ be one of those basis vectors. Send $v$ to 1 and the rest to 0.

But now the restriction of this to $\mathbf{R}$ would be a linear function from $\mathbf{R}$ to $\mathbf{Q}$ other than the 0 function, which does not exist without some amount of choice.

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