$A$ is convex if and only if $\frac{x}{2}+\frac{y}{2} \in A$ for any $x,y \in A$

Question

I'm doing an exercise in a lecture note

Let $A$ be a non-empty set of $\mathbb{R}^{n}$. Show that $A$ is convex if and only if $\frac{x}{2}+\frac{y}{2} \in A$ for any $x,y \in A$.

I feel that this maybe not true. I construct a counter-example as follows. Could you please verify if my understanding is fine?

We define a sequence $(A_n)$ by $A_0 = \{x,y\}$ where $x \neq y$ and $A_{n+1} = \{\frac{a+b}{2} \mid a,b \in A_n\}$. Let $B = \bigcup A_n$. Then $B$ satisfies the condition that $\frac{a+b}{2} \in B$ for all $a,b \in B$. Next we verify that $B$ is not convex. For $b \in B$, we have $b \in A_n$ for some natural number $n$. Then $b = \frac{a_{n-1,1}+a_{n-1,2}}{2} \in B$ for some $a_{n-1,1},a_{n-1,2} \in A_{n-1}$. Recursively, we have $$b = \frac{\sum_{i=1}^{2n} a_{0,i}}{2^n}$$ where $a_{0,i} \in A_0$ for all $i$. As a result, we have for some $k$, $$b = \frac{k}{2^n}x + \frac{(2n-k)}{2^n}y$$

It follows that if $b = \frac{e}{3}x + \frac{3-e}{3} y$ then $b \notin B$.

The set of rational numbers in $\mathbb R$ is a simpler example. — Kavi Rama Murthy, Aug 20 '20 at 08:59
I got it @KaviRamaMurthy. Have you found any mistake in my example? — Akira, Aug 20 '20 at 09:00
Surprized to see that nowhere you use the term "midpoint" for $x/2+y/2$... It is so obvious that a set $K$ is convex iff every midpoint of any segment [a,b] with $a,b \in K$ belongs to $K$... — Jean Marie, Aug 20 '20 at 10:18
I was just meaning that the word "midpoint" is absent, even if you thought to it... But this is a small remark... — Jean Marie, Aug 20 '20 at 10:23

score 2 · Accepted Answer · answered Aug 20 '20 at 09:27

Your example and proof appear to be correct. However, as pointed out in the comments, they are overly complicated. The rational numbers are a much simpler example. If you wanted to modify your own example, notice that it just comes down to numbers with an irrational $x$ coefficient not being in $B$. But you could also notice that your set consists only of rationals anyway, so just take $x=2$ and $y=3$, and note $e\not\in B$.

score 2 · Answer 2 · answered Aug 20 '20 at 09:33

Yes, as mentioned this is false. However, it becomes true when you ask that $A$ is closed and I guess that's what the question asks you to prove.

More generally, if $X$ is a normed space and $A$ is a closed subset of $X$, then convexity of $A$ is equivalent with midpoint convexity of $A$.

$A$ is convex if and only if $\frac{x}{2}+\frac{y}{2} \in A$ for any $x,y \in A$

2 Answers2