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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 $\mathbb{R}=(-\infty,a)\cup[a,+\infty)$)

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Seems like a nice problem! –  Steven Gubkin Apr 23 '10 at 12:14
    
And what about having more pieces in the partition? I guess in this case we don't want them to be intervals. –  Joel David Hamkins Apr 23 '10 at 12:45
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2 Answers 2

up vote 11 down vote accepted

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).

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Nice solution. I had just come up with the same right before you posted. Now my question is: Is AC needed for a solution? –  Steven Gubkin Apr 23 '10 at 12:28
    
I am not an AC fan. Is it possible to show that without AC a desired partition is necessarily trivial? –  Wadim Zudilin Apr 23 '10 at 12:31
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@Wadim: Obviously not, since not assuming AC does not guarantee that AC is not true. You should really ask if “all partitions are trivial” is consistent with ZF without AC. (I don't know the answer.) –  Harald Hanche-Olsen Apr 23 '10 at 12:39
    
It seems you can replace Q in this argument with larger subfields of R and gain even greater degrees of convexity in a nontrivial partition. –  Joel David Hamkins Apr 23 '10 at 12:42
    
Thanks, Harald, for this correction. I am mostly unhappy with the above solution because, in a certain sense, it is "almost trivial": we replace the usual $\mathbb R$-order by the one induced by a $\mathbb Q$-basis... –  Wadim Zudilin Apr 23 '10 at 12:46
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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.

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I think this is close to an answer but I do not understand why A+A containing an interval implies that A. –  Steven Gubkin Apr 23 '10 at 12:47
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The assumptions on $A$ is $(A+A)/2 \subset A$. –  Thomas Kragh Apr 23 '10 at 12:50
    
doh ${ }$ –  Steven Gubkin Apr 23 '10 at 12:52
    
Solovay showed that if the existence of an inaccessible cardinal is consistent with ZFC, then the assertion that every set of reals is Lebesgue measurable is consistent with ZF+DC. So if large cardinals are consistent, then this argument shows that even DC is insufficient to make the desired partition. –  Joel David Hamkins Apr 23 '10 at 12:52
    
Keivan, but don't you need to work a little bit harder? You showed that A or B contains a ray, but you really want that they are both rays. But I think you can just repeat your argument again on the complement of the ray that was found. –  Joel David Hamkins Apr 23 '10 at 12:55
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