Probably someone had asked this question on StackExchange, but can one construct a dense uncountable proper subgroup of $(\mathbb{R},+)$?
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Good question! I don't think the density requirement is really needed. A related question: is it still true if we replace "subgroup" with "subfield"? – goblin GONE Sep 15 '13 at 12:48
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4I think an answer can be found by studying http://math.stackexchange.com/questions/421309/do-proper-dense-subgroups-of-the-real-numbers-have-uncountable-index – Gerry Myerson Sep 15 '13 at 12:49
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Thanks Gerry, so it's not obvious if we don't assume AC... – HYL Sep 15 '13 at 12:56
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If a subgroup of $(\mathbb{R},+)$ is uncountable, it is automatically dense. – Daniel Fischer Sep 15 '13 at 13:00
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2See also this question from MO – Mikko Korhonen Sep 15 '13 at 13:11
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@Hsueh-YungLin Mikko Korhonen's link and Daniel Fischer's comment resolves the problem in ZF. You could wrap them up and answer your own question? – user1729 Sep 15 '13 at 13:55
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This StackExchange question might also be of interest. – Dave L. Renfro Sep 16 '13 at 18:09
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Here's one from Hardy's "A course of pure math":
$\displaystyle \{x \in \mathbb{R}: \lim_{n \to \infty} \sin(n! \pi x) = 0\}$.
hot_queen
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@hot_queen: Could you please give a more precise reference (Edition, Chapter, page perhaps)? Hardy shows there that this set is uncountable? – Jonas Meyer Dec 10 '14 at 22:45
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You can read about it here: http://www.math.wisc.edu/~akumar/BOREL_GROUPS.pdf – hot_queen Dec 10 '14 at 22:46
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Thanks hot_queen. Good example, but it seems a little misleading to say it is from Hardy, at least from the reference given in your link, where the question of what numbers beyond rationals are in the set isn't addressed. Maybe later in the book? – Jonas Meyer Dec 10 '14 at 22:52
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You are right. As far as I remember, Hardy did't consider the question of what reals are in this set beyond the rationals. – hot_queen Dec 10 '14 at 22:55
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Quoting an answer of François Dorais on Mathoverflow:
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.
Jonas Meyer
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