Here's a problem that I've been unable to solve, even though I've tried.
Let $f: \mathbb{S}^{1} \to \mathbb{S}^{1}$ be an orientation-preserving circle homeomorphism with rational rotation number. Let $\Omega(f)$ denote the set of non-wandering points of $f$, and let $Per(f)$ denote the set of periodic points of $f$. Show that $\Omega(f) = Per(f)$.
A point $x \in X$ is non-wandering if for every neighborhood $U$ of $x$, there exists a $k \in \mathbb{N}$ such that $f^{k}(U) \cap U \neq \emptyset$.
It is obvious that $Per(f) \subseteq \Omega(f)$, but I haven't made any substantial progress on proving the other direction.
Note that I tried to use the following known theorem:
If $f \in Homeo^{+}(\mathbb{S}^1)$ and the rotation number of $f$ is rational, then every point $x \in \mathbb{S}^1$ is either periodic or its orbit converges to that of a periodic point, i.e. $\lim_{n\to \infty} d(f^{n}(x), f^{n}(p)) = 0$ for some periodic point $p \in \mathbb{S}^1$.
However, a property that I needed to complete the theorem was equicontinuity of the family $(f^{n})_{n \in \mathbb{N}}$, which I obviously don't have.
What's the best way of approaching this problem? It's possible that I'm on the wrong track completely.