I need some help in proving the following that arises from the proof of the Denjoy's theorem:

Let $f: [0,1] \to [0,1]$ be an orientation preserving circle diffeomorphism topologically conjugated to the rigid rotation $R_f$ with an irrational rotation number $\rho(f)=[a_1, a_2, ...]$.
Set $\frac{p_n}{q_n}=[a_1,a_2,...,a_n]$, prove that $\nexists \ k$ such that: $|k|\le q_n$ and $f^k(x) \in (x, f^{q_n}(x))$ for any $n$.

$\frac{p_n}{q_n}$ is a rational approximation of the rotation number, $(x, f^{q_n}(x))$ is the arc of the circle with shortest length and $[a_1,a_2,...,a_n]$ is the continued fraction representation of the number: $$\frac{1}{a_1+\frac{1}{a_2+\frac{1}{\ddots +\frac{1}{a_n}}}}$$

I've been thinking about it for a while but I don't even know where to start, thank you

  • $\begingroup$ You said it: "$(x,f^{q_n}(x))$ is the arc of the circle with shortest length". Assuming that you can use this, what would happen if you could find such a $k$? Otherwise, your claim is what you should show. $\endgroup$ – John B Dec 9 '16 at 12:51
  • $\begingroup$ I think that the answer to the question is related to the proof, because maybe you can prove this proposition by contradiction showing that the existence of such $k$ mean that exists a better approximation for the irrational rotation number or something similar. $\endgroup$ – pray Dec 9 '16 at 15:36
  • $\begingroup$ As a complement: it is a consequence of the fact that irrational rotations have dense orbits that there are infinitely many integers $q_n$ as you ask, since an orientation-preserving homeomorphism has orbits ordered exactly as those of the rotation with the same rotation number. This is all we need for the proof of Denjoy's theorem, that is, one does not need to know that the numbers $q_n$ come from the continued fraction. $\endgroup$ – John B Dec 9 '16 at 16:45
  • $\begingroup$ I see, thank you. Maybe the formulation of the problem is a bit silly then.. But still I need to prove it for that particular choice of $q_n$ $\endgroup$ – pray Dec 9 '16 at 17:26
  • $\begingroup$ OK, I have added a proof below. $\endgroup$ – John B Dec 9 '16 at 21:47

Claim: There are infinitely many integers $n>0$ such that the intervals $I=(x,f^{-n}(x))$, $f(I), \ldots, f^n(I)$ are pairwise disjoint.

For $k=0,\ldots,n$ we have $$ f^k(I)=\bigl(f^k(x),f^{k-n}(x)\bigr) $$ since $f$ is orientation-preserving. Note that the intervals $f^k(I)$ are pairwise disjoint if and only if $f^k(x) \not \in I$ for $\lvert k \rvert \le n$. But this property only depends on the ordering of the orbit of $x$, which is the same as the ordering of the orbits of the rotation $R$ with the same rotation number (this simpler property is usually shown earlier). Since $\rho(f)$ is irrational, all negative semiorbits are dense. Hence, there exist infinitely many integers $n>0$ such that $R^k(y)\not\in(y,R^{-n}(y))$ for $|k|\le n$ and $y$. Since orbits are ordered in the same manner, this is equivalent to $f^k(x)\not\in(x,f^{-n}(x))$ for $|k|\le n$ and $x$. We are done.


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