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 ?
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asked Dec 28, 2016 at 9:22
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2 Answers
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The problem is generally open. However, recently Liuzhen Wu, Liang Yu, Ralf Schindler and Mariam Beriashvili posted a preprint in which they prove the consistency of the existence of a Hamel basis and $\Bbb R$ cannot be well-ordered. Specifically, they show there is such a basis in Cohen's first model.
This can be found on Ralf's homepage.
answered Dec 28, 2016 at 10:11
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$\begingroup$ But then what do you mean "generally open" ? Is it simply that they didn't publish yet ? $\endgroup$ Commented Dec 28, 2016 at 10:31
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1$\begingroup$ It means that this hasn't been vetted through the usual channels. I read the proof myself, and it looks okay (although I am certain it can be somewhat simplified, at least conceptually, if one takes the approach of symmetric extensions... but then again, I am fluent in the language of symmetric extensions, and less in the language of relative definability.) $\endgroup$– Asaf Karagila ♦Commented Dec 28, 2016 at 11:31
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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
