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 $J_n$? Or a least such that Lebesgue a.e. point has this property?.

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 $J_n$?

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 intervals on 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 $J_n$? Or a least such that Lebesgue a.e. point has this property?.

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 $J_n$? Or a least such that Lebesgue a.e. point has this property?.