First visit of intervals for an irrational rotation

Question

I suppose that what I look for is known, but I can't find it.

Let $\left\lbrace I_n=[a_n,b_n)\right\rbrace$ and $\left\lbrace J_n=[b_n,c]\right\rbrace$ ($n\in\mathbb{N}$) be two countable families of intervals in the unit circle $S^1$. Notice that $I_n$ and $J_n$ are adjacent for every $n$, and that the extremum $c$ is fixed. Assume that the total length of every pair goes to 0, that is $$\lim_{n\to\infty}|c-a_n|= 0.$$ Assume also that the length of the "left" interval is a higher order infinitesimal than the length of the "right" one, that is $$\lim_{n\to\infty}\frac{|b_n-a_n|}{|c-b_n|}= 0.$$

Now let $R:S^1\to S^1$ be an irrational rotation. I say that $x\in S^1$ ultimately first visits $J_n$ if, for every sufficiently large $n$, $$\min\left\lbrace k:R^k(x)\in J_n\right\rbrace<\min\left\lbrace k:R^k(x)\in I_n\right\rbrace.$$ Now my question: is it possible to take the families $\{I_n\}$ and $\{J_n\}$ such that every point in $S^1$ ultimately first visits $J_n$?

Answer 1

No, it is not possible. In the following I will use $I_n=(a_n,b_n)$ instead of $[a_n,b_n)$ (this is not a problem, you can just increase $a_n$ a bit so that the statement with $I_n=(a_n,b_n)$ is stronger).

For fixed $n$, we say that $x$ first visits $J_n$ if $\min\left\lbrace k:R^k(x)\in J_n\right\rbrace<\min\left\lbrace k:R^k(x)\in I_n\right\rbrace$. So the set of points that first visit $J_n$ is $A_n:=\bigcup_{m=1}^\infty\left(R^{-m}J_n\setminus\bigcup_{k=1}^{m-1}R^{-k}I_n\right)$.

Then the set of points that ultimately first visit $J_n$ is $X:=\bigcup_{N=1}^\infty\bigcap_{n=N}^\infty A_n$. What we want is $X=\mathbb{S}^1$, or equivalently, $\varnothing=\mathbb{S}^1\setminus X=\bigcap_{N=1}^\infty\bigcup_{n=N}^\infty(\mathbb{S}^1\setminus A_n)$. Moreover, $\mathbb{S}^1\setminus A_n$ is the set of points that first visit $I_n$, that is, $B_n:=\bigcup_{m=1}^\infty\left(R^{-m}I_n\setminus\bigcup_{k=1}^{m-1}R^{-k}J_n\right)$.

So, we want $\bigcap_{N=1}^\infty\bigcup_{n=N}^\infty B_n=\varnothing$. However, for each $N$, the set $\bigcup_{n=N}^\infty B_n$ is open and dense: it is open because $B_n$ is open for all $n$. To check that it is dense, we will prove that for every $\varepsilon>0$ there is some $n$ such that $B_n$ is $\varepsilon$-dense in $\mathbb{S}^1$. Indeed, let $z$ be the complex number such that $R(x)=zx$ for all $x$. Now let $k$ be such that $\{z,z^2,\dots,z^{k}\}$ is $\varepsilon$-dense in $\mathbb{S}^1$, and let $n$ be so big that the distance between any two points of $\{z,z^2,\dots,z^{k}\}$ is bigger than $|c-a_n|$. This implies that $R^{-a}I_n$ does not intersect $R^{-b}J_n$ if $a,b<k$ (this is obvious if $a=b$, and if not we use that $d(z^{-a},z^{-b})>|c-a_n|$, so as $I_n$ and $J_n$ are contained in an interval of length $|c-a_n|$, $z^{-a}I_n$ doesn't intersect $z^{-b}J_n$). So $B_n$ contains $\bigcup_{m=1}^k\left(R^{-m}I_n\setminus\bigcup_{k=1}^{m-1}R^{-k}J_n\right)=\bigcup_{m=1}^kR^{-m}I_n$, which is $\varepsilon$-dense in $\mathbb{S}^1$ as we wanted.

So as $\bigcup_{n=N}^\infty B_n$ is open and dense for all $N$, by Baire's theorem $\bigcap_{N=1}^\infty\bigcup_{n=N}^\infty B_n$ is nonempty.