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Just what it says on the tin. Let $G$ be a dense subgroup of $\mathbb{R}$; assume that $G \neq \mathbb{R}$. I know that the index of $G$ in $\mathbb{R}$ has to be infinite (since any subgroup of $\mathbb{C}$ of finite index in $\mathbb{C}$ has to have index 1 or 2); does it have to be uncountable, though? All the examples I can readily come up with (e.g. $\mathbb{Q}$, $\mathbb{Z}[\sqrt{2}]$, ...) have uncountable index in $\mathbb{R}$.

Thanks!

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  • $\begingroup$ I think this should work: Let $(\xi_i)$ be a basis for $\mathbb R$ as a vector space over $\mathbb Q$. Now what happens when you remove one element from the set of $\xi_i$? $\endgroup$ – kahen Jun 15 '13 at 18:19
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No: $\mathbb{R}$ is a $\mathbb{Q}$-vector space. Choose a basis and drop finitely many basis elements to find a $\mathbb{Q}$-subspace $G$ of finite codimension. Clearly, $G$ has countable index in $\mathbb{R}$ and is dense.

Added: It can't be done without some form of the axiom of choice. The statement $\mathbb{R} \cong \mathbb{R} \oplus \mathbb{Q}$ is form number 252 in the book Consequences of the Axiom of Choice, by Howard and Rubin, as you can check on the homepage of the book.

C. J. Ash explains in A consequence of the axiom of choice, Journal of the Australian Mathematical Society (Series A), 19 (1975) 306-308, that $\mathbb{R} \cong \mathbb{R} \oplus \mathbb{Q}$ implies the existence of non-measurable sets in $\mathbb{R}$. This in turn is known not to be provable from ZF alone.

If you want to know more about this, you can consult Herrlich, The Axiom of Choice, chapter 5, especially the pages following Diagram 5.10.

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  • $\begingroup$ Great; thanks! One more question: can this be done constructively (that is, w/o axiom of choice)? $\endgroup$ – neilme Jun 15 '13 at 19:17
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    $\begingroup$ I feared that this follow-up question would come :-) See edit. $\endgroup$ – Martin Jun 15 '13 at 19:38
  • $\begingroup$ Excellent; this is all exactly what I wanted to know (and pretty much what I expected, but I really appreciate having the references). Thanks again! $\endgroup$ – neilme Jun 15 '13 at 21:58

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