Prove that if a set is midpoint convex and open, then it is convex

Question

I know that if a set is midpoint convex and closed, then it is convex. Now I want to prove that if $C$ is a midpoint convex and open set, then it is convex. But I don't know how to do it. Could anyone help me to prove it? Every hint is appreciated.

Answer 1

Pick $a,b\in C$ and let $U=\{\,t\in[0,1]\mid ta+(1-t)b\in C\,\}$, $T=[0,1]\setminus U$. As $C$ is open, $U$ is a (relatively) open subset of $[0,1]$, hence $T$ is compact. And clearly, $0\in U$, $1\in U$.

I claim that if $T\ne \emptyset$, then $T$ contains an open interval. So suppose $t_0\in T$ for some $t_0\in [0,1]$. Consider $h$ with $0<h<\min\{t_0,1-t_0\}$ (recall that we must have $0<t_0<1$). If $(t_0-h,t_0+h)\subset T$, we are done. Hence assume $u_0\in U$ for some $u_0\in (t_0-h,t_0+h)$. Then for some $\epsilon>0$, $(u_0-\epsilon,u_0+\epsilon)\subset U\cap (t_0-h,t_0+h)$. By the midpoint property, we conclude that this interval reflected at $t_0$, i.e., $(2t_0-u_0-\epsilon,2t_0-u_0+\epsilon)$ is disjoint from $U$, i.e., is a subset of $T$, as desired.

Now assume $T$ contains an open interval and let $t_1$ be a point inside such an interval. Let $s_1=\sup(U\cap [0,t_1])$ and $s_2=\inf(U\cap [t_1,1])$. Note that $0<s_1< t_1< s_2<1$ and that $(s_1,s_2)\subset T$. Given $\epsilon>0$, we find $u_1\in(s_1-\epsilon,s_1)\cap U$ and $u_2\in (s_2,s_2+\epsilon)$. By the midpoint property, $u^*:=\frac{u_1+u_2}2\in U$. Note that this differs from $\frac{s_1+s_2}2$ by less than $\epsilon$. Hence for $\epsilon<\frac{s_2-s_1}2$, we also have $u^*\in T$, contradiction.

We conclude that $T=\emptyset$.