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The set of rational numbers in $[0,1]$ is used as the counterexample that shows not every midpoint convex set is convex. To see the fact notice that the set of rational numbers in $[0,1]$ is midpoint convex since for any pair of rational numbers $x,y$ in $[0,1]$ $\frac{x+y}{2}$ is a rational number in $[0,1]$.

Now the question is how to show the set of rational numbers in $[0,1]$ is not convex.

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  • $\begingroup$ @herb steinberg: can you please write this as an answer so that I can approve it. It would be awesome to add an example. $\endgroup$
    – Saeed
    Jun 26, 2021 at 21:14
  • $\begingroup$ @amWhy: thank you for your comment and giving feedback to members. I appreciate your time. I edited the title and tried to modify the question statement. Hope my modifications has made the posted question better. $\endgroup$
    – Saeed
    Jun 27, 2021 at 0:42
  • $\begingroup$ Thanks for your edit to the title. I think it helps improve your post. Take care! $\endgroup$
    – amWhy
    Jun 27, 2021 at 0:44

1 Answer 1

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Use weights $z$ and $1-z$ with $0< z < 1$ and $z$ irrational.

Moreover, the set of irrational is not convex, which is more tricky. Consider the following counterexample:

$$za+(1-z)b$$ for $$z=\frac{1}{\sqrt{2}}, a=\frac{1}{\sqrt{2}}, b=\frac{1}{2}\left ( 1-\frac{1}{\sqrt{2}} \right).$$

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  • $\begingroup$ Why -1???????????? $\endgroup$ Jun 27, 2021 at 15:21
  • $\begingroup$ The answer is correct. I just improved and upvoted. $\endgroup$
    – Amir
    Dec 31, 2023 at 14:37

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