# For any orientation-preserving circle homeomorphism $f$ with rational rotation number , $\Omega(f) = Per(f)$

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.

• Do you mean with rational rotation number? Otherwise, the statement is false. For rational rotation number, your "known theorem" gives it immediately since it says that nonperiodic points cannot be nonwandering (think of taking multiples of the period). Anyways, you don't need equicontinuity of the family (even though you clearly have it). – John B Mar 15 at 2:53
• Yes, the rotation number is rational. I'm sorry I didn't put that in the question. I edited it now. Anyway, could you maybe explain in detail why the theorem I stated immediately gives the answer? I couldn't come up with one, even though it might be easy to you. – Matija Sreckovic Mar 15 at 7:01
• Also, could you explain why the family $(f_{n})_{n \in \mathbb{N}}$ is equicontinuous? – Matija Sreckovic Mar 15 at 7:04
• My problem was that the "non-wandering-ness" of $x$ doesn't necessarily give me anything about the "location" of $f^{n}(x)$, only about $f^{n}$ of some point close to $x$. And if I take a sequence of $U$'s whose diamater tends to zero, I get a sequence $x_{n} \to x$, with $f^{k_{n}}(x_{n}) \to x$, where $k_{n}$ is a sequence of natural numbers that I don't know much about, which is why I wanted equicontinuity. – Matija Sreckovic Mar 15 at 7:25
• I suggest that you make a drawing in the circle with such a point $p$ as in the "known theorem" and then think of what happens when you iterate with multiples of the period. Hince: since $f$ is a homeomorphism, it must be increasing or decreasing outside periodic points. – John B Mar 15 at 11:47