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It’s well-known that the AC implies that$\mathbb{R}$ is a $\mathbb{Q}$-vector space of infinite dimension, and that the cardinality of a Hamel base of that space is $2^{\aleph_{0}}$. Is there any algebraic or analytical method to give a explicit base? If there isn’t any, why not? I’ve been thinking of a method involving the equivalence relation $a\sim b \iff \frac{a}{b} \in \mathbb{Q}$ in $\mathbb{R}^{\times}$ and choosing a representative of each equivalence class of $\mathbb{R}^{\times}/\sim$. One of the problems with this method is that the Axiom of Choice is needed, so I was wondering whether there is a different method which doesn’t need it (if it’s also needed to proof its existence, why? Since if the AC is needed to proof the existence of a basis, it must be needed to construct a explicit one).

JuanClaver
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    You wrote It's well known..., but the thing that is well known is not the thing that you wrote. Instead, what is well known is that the axiom of choice implies that the cardinality of a Hamel base of that space is $2^{\aleph_{0}}$. – Lee Mosher Sep 06 '24 at 13:21
  • @LeeMosher you're right. I should have added "Assuming the axiom of choice as true..." – JuanClaver Sep 07 '24 at 12:38

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It is consistent with the negation of choice that there is no Hamel basis for $\mathbb{R}$. Indeed, the existence of such a basis implies the existence of a non-measurable set (this is the standard example of a non-measurable set, constructed by picking a member of every equivalence class of $\mathbb{R}/\mathbb{Q}$ as you say), but it's consistent with the negation of choice that all sets are Lebesgue measurable.

hunter
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