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
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.