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Unlike the case with bases, AC is not needed to prove (even constructively) the existence of uncountable linearly independent subsets of $\mathbb{R}$ considered as a $\mathbb{Q}$-vector space, see e.g. https://mathoverflow.net/questions/23202/explicit-big-linearly-independent-sets

What if I'm not even interested in an explicit construction, but only in a merely existential proof (without AC), and I don't even want size continuum, but mere uncountability.

What's the simplest argument one can give?

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Damian Reding
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  • That I don't quite understand. The definition of linear independence does not need a basis. Therefore the existence of an uncountable linearly independent set should be independent of the existence of a basis. Though, lacking a generally accepted norm for "simplicity", this is quite an opinion based question. Also, the construction in the link is rather simple... – senegrom Dec 01 '16 at 15:35
  • @senegrom I misread. You're right. – user1551 Dec 01 '16 at 15:42

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