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 $X$. 


My question is: is it possible to partition $\mathbb{R}$ into two midpoint convex sets in a non-trivial way? 


(trivial way is $\mathbb{R}=(-\infty,a]\cup(a,+\infty)$ or $\mathbb{R}=(-\infty,a)\cup[a,+\infty)$) 



Partition of R into midpoint convex sets
Yes if you assume AC:


With AC let $\{v\_\alpha\}$ be a $\mathbb{Q}$-basis for $\mathbb{R}$ then the following two sets satisfies your property:


$A = \{q\_1v\_{\alpha\_1}+\cdots+q\_nv\_{\alpha\_n} \mid q\_i \in \mathbb{Q} , \sum q\_i \geq 0 \}$


and 


$B = \{q\_1v\_{\alpha\_1}+\cdots+q\_nv\_{\alpha\_n} \mid q\_i \in \mathbb{Q} , \sum q\_i < 0 \}$


So in fact these are $\mathbb{Q}$ convex (in the obvious sense).


