most recent 30 from http://mathoverflow.net 2013-05-23T19:57:25Z http://mathoverflow.net/feeds/question/59978 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/59978/additive-subgroups-of-the-reals Alex 2011-03-29T15:39:21Z 2012-04-12T14:48:09Z

<p>Does anyone know if there is a classification of the subgroups of the real numbers taken under addition? If not can anyone point me in the directiong of any papers/materials which discuss properties of or interesting facts about these subgroups? </p>

http://mathoverflow.net/questions/59978/additive-subgroups-of-the-reals/59980#59980 Emil Jeřábek 2011-03-29T15:51:56Z 2011-03-29T16:19:20Z

<p>Every torsion-free abelian group of cardinality at most $2^\omega$ is isomorphic to a subgroup of the reals. (To see this, note that any such group can be embedded in a divisible torsion-free group of the same cardinality, i.e., a vector space over $\mathbb Q$, which can in turn be embedded in any other vector space over $\mathbb Q$ of the same or greater dimension.) Since already the structure of <a href="http://en.wikipedia.org/wiki/Rank_of_an_abelian_group" rel="nofollow">rank</a> 2 abelian groups is hopelessly complicated, you are not going to find any sensible classification.</p>

http://mathoverflow.net/questions/59978/additive-subgroups-of-the-reals/59984#59984 Ashutosh 2011-03-29T16:40:26Z 2011-03-29T16:40:26Z

<p>If you're interested in topological classification, then this might be useful: Farah and Solecki - Borel subgroups of Polish groups, Advances in Mathematics 199, 2006, 499-541.</p> <p>Among a lot of other things, one of their results shows that for any countable ordinals $\alpha \neq 2$ and $\beta \geq 2$, there are $\Pi_{\alpha}^{0}$-complete and $\Sigma_{\beta}^{0}$-complete additive subgroups of any uncountable polish group. For connected abelian polish groups, this was previously shown by Mauldin by refining a result of Klee.</p>

http://mathoverflow.net/questions/59978/additive-subgroups-of-the-reals/59985#59985 Juris Steprans 2011-03-29T16:49:04Z 2011-03-29T16:49:04Z

<p>The 2002 paper of Simon Thomas in JAMS provides a precise measure of "hopelessly complicated". As the rank of the groups increases, so does the complexity of the classification problem.</p>

http://mathoverflow.net/questions/59978/additive-subgroups-of-the-reals/59988#59988 Syang Chen 2011-03-29T17:00:52Z 2011-03-29T17:28:25Z

<p>If you would like to classify the subgroups in the sense of Lebesugue measure, you may find the following facts helpful.</p> <p>(1) Any measurable proper subgroup of the real line is of measure $0$.</p> <p>(2) Any non-measurable subgroup $G$ of the real line charges fully everywhere, i.e., for any interval $I$, $m^{\ast}(G \cap I)=|I|$, where $m^{\ast}(\cdot)$ denotes the outer Lebesgue measure.</p> <p>(3) Non-measurable subgroup of the real line exists.</p>

http://mathoverflow.net/questions/59978/additive-subgroups-of-the-reals/59992#59992 Thierry Zell 2011-03-29T17:19:11Z 2011-03-29T17:19:11Z

<p>I'm surprised no one has yet stated the most obvious fact (though I guess Xianghong's answer comes pretty close), which is that an additive subgroup of the reals is either of the form $a\mathbb{Z}$ or is dense in the real line (an obvious consequence from division with remainder).</p>

http://mathoverflow.net/questions/59978/additive-subgroups-of-the-reals/93870#93870 Alain Valette 2012-04-12T14:48:09Z 2012-04-12T14:48:09Z

<p>For every real number $\alpha$ with $0&lt;\alpha&lt;1$, there is an uncountable subgroup of $\mathbb{R}$ with Hausdorff dimension $\alpha$: see e.g. <a href="https://perswww.kuleuven.be/~u0018768/artikels/actions-free-group.pdf" rel="nofollow">https://perswww.kuleuven.be/~u0018768/artikels/actions-free-group.pdf</a></p>