I came up with the following exercise. Let $X \subset \mathbb{R}$ be a set containing non-negative numbers satisfying the following properties:
- $0 < m= \sup X < 1 $
- for every $x,y \in X$ s.t $x < y$, $xy^{-1} \in X$
prove that $m= \max X$.
A possible solution is: suppose $m \notin X$, then there exists a strictly monotone sequence $(x_n)_n \subset X$ such that $x_n \rightarrow m$; thus $(x_n x_{n+1}^{-1})_n$ is a sequence of points of $X$ (by the second hypothesis) converging to $1$. This easily yields an absurd conclusion since we can find points of $X$ in the latter sequence arbitrary close to $1$ but $\sup X < 1$.
Thinking a bit more on such a set, I realized that it should have a "countable nature": it cannot be an interval and moreover the only examples I could find are sets like: $$ X=\{ 3^{-n} : n \ge 1 \} $$
With this in mind, I was wondering about the existence of a different, more "discrete" and less analytical proof. For instance, I thought to consider the set:
$$ S=\{ x \in X : xm^{-1} \notin X \} $$
Again supposing $m \notin X$, $S$ turns out to be non empty (in fact, we know a posteriori that $S$ contains only the singleton $m$). Unfortunately I am not able to continue on this path and conclude in an elegant way: maybe there is simply no proof radically different from the first one I gave. I would be pleased to see some ideas or insights on this one.