I need to show that if $f:S^{1} \rightarrow S^{1}$ is a preserving-order diffeomorphism and $f$ has irrational rotation number, then $f$ has at least one recurrent point. How can I prove that?

Definition: $x_0$ is a recurrent point under $f$ if, for any neighborhood $U$ of $x_0$, there exists $n>0$ such that $f^{n}(x_0) \in U $.

*It is known that a preserving-order diffeomorphism has irrational rotation number if and only if it has no periodic points.

**Maybe the hypothesis of preserving-order diffeomorphism is too restrict, and we only need homeomorphism.

  • $\begingroup$ Every continuous dynamical system on a compact metric space has a recurrent point, so it is certainly true in much more general cases. $\endgroup$ – Lukas Geyer Jun 3 '15 at 17:09
  • $\begingroup$ @LukasGeyer But that is what I want to prove... $\endgroup$ – mathgccunha Jun 3 '15 at 17:29

This is a corollary of a more general result, that every continuous self-map $f:X \to X$ of a (non-empty) compact metric space $X$ has a recurrent point. Here is a proof, stolen from Katok / Hasselblatt, Modern Theory of Dynamical Systems, Cambridge University Press 1995, Proposition 3.3.6 and Corollary 3.3.7.

Let $\mathcal{C}$ be the collection of all non-empty compact subsets $A \subseteq X$ with the property that $f(A) \subseteq A$, partially ordered by inclusion. If $(A_i)_{i \in \mathcal{I}}$ is a non-empty chain in $\mathcal{C}$, i.e., a totally ordered subset, then the intersection $A = \bigcap_{i \in \mathcal{I}} A_i$ is again non-empty and compact and satisfies $f(A) \subseteq A$, so $A \in \mathcal{C}$. By Zorn's lemma $\mathcal{C}$ has a minimal element $A^*$. Pick $x \in A^*$ and let $A_x$ be the closure of the forward orbit of $x$. Then $A_x \subseteq A^*$ and $A_x \in \mathcal{C}$, so $A_x = A^*$, and in particular $x$ is recurrent.


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