Partition of R into midpoint convex sets - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T13:10:26Z http://mathoverflow.net/feeds/question/22327 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/22327/partition-of-r-into-midpoint-convex-sets Partition of R into midpoint convex sets filipm 2010-04-23T10:34:05Z 2010-04-23T12:41:42Z <p>We say that a subset $X$ of $\mathbb{R}$ is midpoint convex if for any two points $a,b\in X$ the midpoint $\frac{a+b}{2}$ also lies in $X$. </p> <p>My question is: is it possible to partition $\mathbb{R}$ into two midpoint convex sets in a non-trivial way? </p> <p>(trivial way is $\mathbb{R}=(-\infty,a]\cup(a,+\infty)$ or $\mathbb{R}=(-\infty,a)\cup[a,+\infty)$) </p> http://mathoverflow.net/questions/22327/partition-of-r-into-midpoint-convex-sets/22338#22338 Answer by Thomas Kragh for Partition of R into midpoint convex sets Thomas Kragh 2010-04-23T12:18:43Z 2010-04-23T12:18:43Z <p>Yes if you assume AC:</p> <p>With AC let $\{v_\alpha\}$ be a $\mathbb{Q}$-basis for $\mathbb{R}$ then the following two sets satisfies your property:</p> <p>$A = \{q_1v_{\alpha_1}+\cdots+q_nv_{\alpha_n} \mid q_i \in \mathbb{Q} , \sum q_i \geq 0 \}$</p> <p>and </p> <p>$B = \{q_1v_{\alpha_1}+\cdots+q_nv_{\alpha_n} \mid q_i \in \mathbb{Q} , \sum q_i &lt; 0 \}$</p> <p>So in fact these are $\mathbb{Q}$ convex (in the obvious sense).</p> http://mathoverflow.net/questions/22327/partition-of-r-into-midpoint-convex-sets/22342#22342 Answer by Keivan Karai for Partition of R into midpoint convex sets Keivan Karai 2010-04-23T12:41:42Z 2010-04-23T12:41:42Z <p>well, if we assume that $A$ and $B$ are measurable then at least one of them (say A) should have positive measure, and since A+A for a set of positive measure contains an interval, A contains and interval, say $(a,b)$. Then I think it is easy to show that for the maximal interval of this type $a$ or $b$ must be infinity, since otherwise, by taking two sequences in $B$ converging from different sides, you get a contradiction. Solovey has constructed models of ZF in which every set is measurable, so I think this implies you need something like AC.</p>