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With the axiom of choice, it is easy to show that there is an uncountable proper subgroup of the additive group of real numbers. Namely, if $H$ is a Hamel basis of $\mathbb{R}$ over $\mathbb{Q}$, then for any $x \in H$, the $\mathbb{Q}$-vector subspace of $\mathbb{R}$ generated (or spanned) by $H \setminus \{x\}$ is an uncountable proper subgroup.

But, could an uncountable proper subgroup of the additive group of real numbers still be proven to exist without the axiom of choice?

Geoffrey Trang
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    Yes: in fact, it's possible to devise $\Bbb Q$-linearly independent subsets of $\Bbb R$ of cardinality $\beth_1$. See this and this on MO. In the first one, one of the answers suggests that there might even be subsets of $\Bbb R$ of cardinality $\beth_1$ that are provably algebraically independent in ZF (I did not check the articles, though). –  Nov 19 '21 at 23:14
  • @SaucyO'Path Re: definability, in fact we can do even better: there is a single formula which $\mathsf{ZF}$ proves defines a size-continuum set of reals $C$ no finite subset of which computes any other, which for examplies implies algebraic independency. – Noah Schweber Nov 20 '21 at 00:28

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