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It is straightforward to construct proper uncountable subgroups of $\mathbb{R}$. One can construst a basis for $\mathbb{R}$ over $\mathbb{Q}$, and then there are many possibilities (just consider the group generated by the basis or the vector subspace generated by some proper uncountable set of the basis).

However, the first step (constructing the basis) requires the axiom of choice.

So does anyone know of any proper uncountable subgroup of $\mathbb{R}$ that does not require choice to construct?

or is this not possible.

Meaning are there models not involving choice where every uncountable subgroup of $\mathbb{R}$ is equal to $\mathbb{R}$.

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3 Answers 3

up vote 15 down vote accepted

This earlier answer of mine shows how to get an uncountable $\mathbb{Q}$-independent subset of $\mathbb{R}$ in ZF. This set is not a Hamel basis so the $\mathbb{Q}$-span of this set is as required.

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+1.. Very nice! –  Malik Younsi Aug 18 '10 at 18:53

For any subset $S$ of the positive integers, let $\alpha_S =\sum_{i\in S} 10^{-i!}$. Then it's not hard to show (if I haven't made a mistake) that the subgroup of $\mathbb{R}$ generated by the $\alpha_S$'s is uncountable and proper.

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There's a Borel example of such groups here

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François' example is also Borel, since it is arithmetically definable. –  Joel David Hamkins Aug 18 '10 at 22:07

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