TITLE: Relation between well-orderings of $\mathbb{R}$, and bases over $\mathbb{Q}$
QUESTION [8 upvotes]: The following question arose from a discussion about the definability of bases of $\mathbb{R}$ as a $\mathbb{Q}$-vector space. 
(ZF without AC) something we can note is that the existence of a (definable) well-ordering of $\mathbb{R}$ is easily seen to be equivalent to that of a (definable) well-ordered basis of $\mathbb{R}$ as a $\mathbb{Q}$-vector space.
After these remarks, the following question seems natural : 
Is it consistent with ZF that there be a basis of $\mathbb{R}$ over $\mathbb{Q}$ that cannot be well-ordered ?

REPLY [6 votes]: There are actually two papers: "ZF + there is a Hamel basis" doesn't give that every infinite set of reals has a ctble. subset, see  https://ivv5hpp.uni-muenster.de/u/rds/hamel_basis.pdf , and "ZF + DC + there is a Hamel basis" doesn't give that there is a w.o. of R, see  https://ivv5hpp.uni-muenster.de/u/rds/hamel_basis_2.pdf