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It is consistent with the axioms of ZFC that there is a Hamel basis of $\mathbb{R}$ over $\mathbb{Q}$ with complexity $\Delta^1_2$ in the descriptive set theoretic hierarchy. This is true, for example, in the constructible universe $L$, where there is a $\Delta^1_2$ well-ordering of the reals, as I explain in this MO answer, which is closely related to this question. Complexity $\Delta^1_2$ is a surprisingly low complexity, since such a set (and its complement) can both be obtained by starting with a certain closed set in $\mathbb{R}^3$, projecting it to $\mathbb{R}^2$, taking the complement, and projecting down to $\mathbb{R}$, and so it would seem to count as fairly explicit.

Meanwhile, there can never be a Hamel basis of $\mathbb{R}$ over $\mathbb{Q}$ that is Borel, that is, with complexity $\Delta^1_1$, since from any Borel Hamel basis one can produce a non-Lebesgue measurable set of the same complexity by the Vitali argument (remove an element, take the span of the other elements, and consider its cosets). But of course every Borel set is Lebesgue measurable.

At the same time, it is a consequence of the existence of large cardinals that every projective set of reals is Lebesgue measurable, and in this case, there can be no projective Hamel basis for $\mathbb{R}$ over $\mathbb{Q}$, again by the Vitali argument. The projective hierarchy of sets arises by closing the Borel sets under continuous images, as well as complements, countable unions and intersections. Thus, in such a situation, there can be no easily-described Hamel basis for $\mathbb{R}$ over $\mathbb{Q}$.

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It is consistent with the axioms of ZFC that there is a Hamel basis of $\mathbb{R}$ over $\mathbb{Q}$ with complexity $\Delta^1_2$ in the descriptive set theoretic hierarchy. This is true, for example, in the constructible universe $L$, where there is a $\Delta^1_2$ well-ordering of the reals, as I explain in this MO answer, which is closely related to this question. Complexity $\Delta^1_2$ is a surprisingly low complexity, since such a set (and its complement) can both be obtained by starting with a certain closed set in $\mathbb{R}^3$, projecting it to $\mathbb{R}^2$, taking the complement, and projecting down to $\mathbb{R}$, and so it would seem to count as fairly explicit.

Meanwhile, there can never be a Hamel basis of $\mathbb{R}$ over $\mathbb{Q}$ that is Borel, that is, with complexity $\Delta^1_1$, since from any Hamel basis one can produce a non-Lebesgue measurable set of the same complexity by the Vitali argument (remove an element, take the span of the other elements, and consider its cosets), and if the Hamel basis is Borel, then this set will be analytic $\Sigma^1_1$. . But of course every analytic Borel set is Lebesgue measurable.

At the same time, it is a consequence of the existence of large cardinals that every projective set of reals is Lebesgue measurable, and in this case, there can be no projective Hamel basis for $\mathbb{R}$ over $\mathbb{Q}$, again by the Vitali argument. The projective hierarchy of sets arises by closing the Borel sets under continuous images, as well as complements, countable unions and intersections. Thus, in such a situation, there can be no easily-described Hamel basis for $\mathbb{R}$ over $\mathbb{Q}$.

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It is consistent with the axioms of ZFC that there is a Hamel basis of $\mathbb{R}$ over $\mathbb{Q}$ with complexity $\Delta^1_2$ in the descriptive set theoretic hierarchy. This is true, for example, in the constructible universe $L$, where there is a $\Delta^1_2$ well-ordering of the reals, as I explain in this MO answer, which is closely related to this question. Complexity $\Delta^1_2$ is a surprisingly low complexity, since such a set (and its complement) can both be obtained by starting with a certain closed set in $\mathbb{R}^3$, projecting it to $\mathbb{R}^2$, taking the complement, and projecting down to $\mathbb{R}$, and so it would seem to count as fairly explicit.

Meanwhile, there can never be a Hamel basis of $\mathbb{R}$ over $\mathbb{Q}$ that is Borel, that is, with complexity $\Delta^1_1$, since from any Hamel basis one can produce a non-Lebesgue measurable set of the same complexity by the Vitali argument (remove an element, take the span of the other elements, and consider its cosets). , and if the Hamel basis is Borel, then this set will be analytic $\Sigma^1_1$. But of course every Borel analytic set is Lebesgue measurable.

At the same time, it is a consequence of the existence of large cardinals that every projective set of reals is Lebesgue measurable, and in this case, there can be no projective Hamel basis for $\mathbb{R}$ over $\mathbb{Q}$, again by the Vitali argument. The projective hierarchy of sets arises by closing the Borel sets under continuous images, as well as complements, countable unions and intersections. Thus, in such a situation, there can be no easily-described Hamel basis for $\mathbb{R}$ over $\mathbb{Q}$.

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