Closed Under Means

Here's a nice little problem from the 13th All Soviet Union Mathematics Olympiad.

Given a set of real numbers $S$ containing 0 and 1 that's closed under finite means, show that it contains all rational numbers in the interval $\left[0,1\right]$.

This isn't a difficult problem, but there's a particularly nice solution.

First observe that if $x\in S$ then both $\frac{x}{4}$ and $\frac{3x}{4}$ are in $S$; average $\{0,x\}$ to get $\frac{x}{2}$, average $\{0, \frac{x}{2}\}$ to get $\frac{x}{4}$, average $\{\frac{x}{2}, x\}$ to get $\frac{3x}{4}$.

We could show any rational number $\frac{m}{n}$ with $1\leq m < n$ is in $S$ if we had $n$ distinct elements from $S$ that summed to $m$. Let's exhibit one.

Start with an array with $m$ 1s on the left, $n-m$ 0s on the right. Repeatedly replace adjacent $x,y$ values with $\frac{3(x+y)}{4}, \frac{(x+y)}{4}$, where either $x=1,y\neq1$ or $x\neq 0, y=0$, until there is one 0 and one 1 left. We can do this in exactly $n-2$ steps, each resulting number is guaranteed to be in $S$ by the above note, and each number is guaranteed to be distinct by construction!

Examples:

$\frac{1}{3}: \left[1,0,0\right] \to \left[\frac{3}{4},\frac{1}{4},0\right]$

$\frac{2}{5}: \left[1,1,0,0,0\right] \to \left[1,\frac{3}{4},\frac{1}{4},0,0\right] \to \left[1,\frac{3}{4},\frac{3}{16},\frac{1}{16},0\right]$