A closed set is convex only when $(x+y)/2$ is in the set for all $x,y$

Question

How can one prove that:

Given that $A$ is a closed subset of $\mathbb{R}^{n}$, $A$ is convex $\iff \frac{1}{2 }(x+y) \in A$, $\forall x,y \in A.$

I know $\frac{1}{2}(x+y)$ is a extreme point, but I 'm not sure how to link it with the property of closed set. Thank you in advance.

No, if $x \ne y$ then $(x+y)/2$ is not an extreme point. — Robert Israel, Sep 19 '15 at 01:58
thank you a lot!! I just started to learn this~~~ Can you give me some clue about how to prove the above question — user272101, Sep 19 '15 at 02:15

score 4 · Answer 1 · answered Sep 19 '15 at 02:00

4

Hint: if $x, y \in A$ and $0 \le t \le 1$, approximate $t x + (1-t) y$ by $s x + (1-s) y$ where $s$ is a rational number with denominator a power of $2$.

answered Sep 19 '15 at 02:00

Robert Israel

448,999

oh, I see. I guess was thinking in a wrong way. Thanks!! – user272101 Sep 19 '15 at 02:21

A closed set is convex only when $(x+y)/2$ is in the set for all $x,y$

1 Answers1