Construction of a proper uncountable subgroup of $\mathbb{R}$ without Choice. - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T07:49:54Z http://mathoverflow.net/feeds/question/35970 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/35970/construction-of-a-proper-uncountable-subgroup-of-mathbbr-without-choice Construction of a proper uncountable subgroup of $\mathbb{R}$ without Choice. Owen Sizemore 2010-08-18T14:03:40Z 2010-08-19T19:02:44Z <p>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). </p> <p>However, the first step (constructing the basis) requires the axiom of choice.</p> <p>So does anyone know of any proper uncountable subgroup of $\mathbb{R}$ that does not require choice to construct?</p> <p>or is this not possible.</p> <p>Meaning are there models not involving choice where every uncountable subgroup of $\mathbb{R}$ is equal to $\mathbb{R}$. </p> http://mathoverflow.net/questions/35970/construction-of-a-proper-uncountable-subgroup-of-mathbbr-without-choice/35974#35974 Answer by François G. Dorais for Construction of a proper uncountable subgroup of $\mathbb{R}$ without Choice. François G. Dorais 2010-08-18T14:54:35Z 2010-08-18T14:54:35Z <p>This <a href="http://mathoverflow.net/questions/23202/explicit-big-linearly-independent-sets/23206#23206" rel="nofollow">earlier answer of mine</a> 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.</p> http://mathoverflow.net/questions/35970/construction-of-a-proper-uncountable-subgroup-of-mathbbr-without-choice/36006#36006 Answer by Ashutosh for Construction of a proper uncountable subgroup of $\mathbb{R}$ without Choice. Ashutosh 2010-08-18T18:43:39Z 2010-08-18T18:43:39Z <p>There's a Borel example of such groups <a href="http://mathoverflow.net/questions/34369/on-the-behaviour-of-sinn-pi-x-when-x-is-irrational/34436#34436" rel="nofollow">here</a></p> http://mathoverflow.net/questions/35970/construction-of-a-proper-uncountable-subgroup-of-mathbbr-without-choice/36113#36113 Answer by Richard Stanley for Construction of a proper uncountable subgroup of $\mathbb{R}$ without Choice. Richard Stanley 2010-08-19T19:02:44Z 2010-08-19T19:02:44Z <p>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.</p>