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 $J\_n$?



First visit of intervals for an irrational rotation
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](https://en.wikipedia.org/wiki/Dense_set#Related_notions) 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.


