Does there exist a compact, nonempty subset of the rationals without isolated points ?
My motivation was the following: If so one could define a map $f$ from the set of all compact subsets of the rationals to itself that sends $K$ to $K\setminus Is(K)$ where $Is(K)$ denotes the set of all isolated points of $K$. Since $Is(K)$ is an open subset of $K$ the result is again compact. Then we can define $f^n$ for any ordinal number $n$ via $f^{n+1}=f\circ f^n$ and via $f^n:=\bigcap_{m\le n} f^m$ in the limit case. If every compact set has an isolated point it follows that for every $K$ there is a (smallest) ordinal $n$ such that $f^n(K)=\emptyset$. So we could measure the complexity of compact subsets of $Q$ with ordinal numbers.