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It can be easily shown that a closed midpoint convex set in $\mathbb{R}^n$ is always convex, but it has occurred to me that the counter-examples showing a midpoint convex set may not be convex, are sets that are neither open nor closed.

Can anyone show me an open midpoint convex set in $\mathbb{R}^n$ that is not convex?

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I presume midpoint convex means the midpoint of two points of the set is also an element of the set.

Let $U$ be an open midpoint convex subset of $\Bbb R^n$. Let $P$, $Q$ be distinct points of $U$. Let $L$ be the line joining $P$ and $Q$. Then $L\cap U$ is an open midpoint convex subset of $L$. Thus we can reduce the problem to proving that an open midpoint convex subset of $\Bbb R$ is convex.

Now let $U\subseteq\Bbb R$ be open and midpoint convex. Let $P$, $Q\in U$ be distinct. We can assume $P=0$, $Q=1$. There is $\newcommand{\ep}{\varepsilon}\newcommand{\sub}{\subseteq}\ep>0$ such that $(-\ep,\ep)\sub U$ and $(1-\ep,1+\ep)\sub U$. By midpoint convexity, $(1/2-\ep,1/2+\ep)\sub U$, $(1/4-\ep,1/4+\ep)\sub U$, $(3/4-\ep,3/4+\ep)\sub U$, etc. Thus $(r-\ep,r+\ep)\sub U$ for all dyadic rationals $r\in[0,1]$. But these intervals cover $[0,1]$. Then $[0,1]\sub U$ and $U$ is convex.

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  • $\begingroup$ Brilliant! Thank you! $\endgroup$
    – Kurosu
    Apr 6 '19 at 7:25

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