If $\frac{1}{2}A + \frac{1}{2}A \subset A$, and $A$ is closed, then $A$ is a convex set

I was just recently presented this very short problem in convex sets:

Suppose that $A \subset R^n$ is a closed set satisfying $\frac{1}{2}A + \frac{1}{2}A \subset A$, we are to show that A is a convex set.

Obviously this can be solved with topology in a brute force sort of way involving A being a closed set. I am not sure about it though, might there be a nice slick proof of this as I cannot really and formally write one myself. Help appreciated thanks

• TFAE: (1) $A$ is closed and convex; (2) $A$ is closed and $\frac12 A + \frac12 A = A$; (3) $A$ is the intersection of some (possibly infinite) number of closed half-spaces. Commented Sep 14, 2015 at 18:02

Iterating the given, we obtain that

$$\frac{1}{2^n} A + \frac{1}{2^n} A + \dots + \frac{1}{2^n}A \subset A$$

where there are $2^n$ summands on the left.

Now, given $\lambda, \mu$ with $\lambda+\mu = 1$, for each $n$ take $k,m$ such that $\frac{k}{2^n}$ and $\frac{m}{2^n}$ are "close" to $\lambda, \mu$ and $k+m = 2^n$.

Then apply the above to conclude that for $x,y \in A$, we have $\frac{k}{2^n}x + \frac{m}{2^n}y \in A$.

Take $n \rightarrow \infty$ and use that $A$ is closed to get the conclusion.

• Note that we really need a closed set $A$, otherwise, the set where all coordinates are rational numbers would satisfy said relation but is not convex. Commented Sep 14, 2015 at 12:55