Liminf and limsup of $A_n = \{m/n: m \in \mathbb N\}$ I'm doing some exercises on Chapter 1 of Resnick's A Probability Path and got confused in this question:

Suppose $A_n = \{m/n: m \in \mathbb N\}, n \in\mathbb N$,
  where $\mathbb N$ are non-negative integers, what are
  $\liminf A_n$ and $\limsup A_n$?

I understand both definition of $\liminf A_n$ and $\limsup A_n$ are:
$$\liminf A_n = \bigcup_{n\ge 1}\bigcap_{k\ge n}A_k\qquad\limsup A_n = \bigcap_{n\ge 1}\bigcup_{k\ge n}A_k$$
I tried to do $\liminf$ but I've got null set just by observation which is of course not recommended when answering mathematical questions.
What confuses me here is $m$ in a way that I don't know how to use this in a sequence since sequences should have a "pattern going somewhere" as $n$ goes to $\infty$. Surely, I can't treat $m$ as something that approaches some integer (i.e. m shouldn't be treated as increasing or decreasing).
 A: $A_n = \{ \frac{m}{n} ,m \in \mathbb{N}\}$
$\limsup A_n  = \{ x | x \in A_n $ for infinitely many $n\}$.
$\liminf A_n = \{ x | x \notin A_n $ only for finitely many $n\}$.
It is clear that no irrational numbers belong to $A_n$ for any $n$. So it is enough for us to consider rational numbers.
$x \in A_n \iff xn=m$ for some $m  \in \mathbb{N}$.
If we write $x=\frac{p}{q}$ in lowest form (It's possible that $q=1$), then $xn=m$ for some $m$ if and only if $n | q$.  That is, $x \in A_{kq} \forall k \in \mathbb{N}$, since $\frac{p(kq)}{q} = pk \in \mathbb{N}$. Thus, every rational number $x$, $x \in \limsup A_n$.This shows that $\limsup A_n = \mathbb{Q}$.
If $x \in \liminf A_n$, then there is some natural number $N$ such that $x \in A_n$ if $n > N$. Now, we have that $\frac{p(N+1)}{q}$ is an integer, and $\frac{p(N+2)}{q}$ is an integer. Subtracting, we have that $\frac{p}{q}$ is an  integer,which happens only when $x$ is an integer itself. On the other hand, of course an integer is in $A_n$ for every $n$ . Hence, it follows that $\liminf A_n = \mathbb{N}$.

I'm keeping my explanation on, since you are attempting to understand it.

$\limsup A_n  =\displaystyle\bigcap_{n=1}^\infty \bigcup_{k \geq n} A_k$.
$\liminf A_n  =\displaystyle\bigcup_{n=1}^\infty \bigcap_{k \geq n} A_k$.

$\forall i \in \mathbb{R}-\mathbb{Q }$, $i \notin A_n \ \forall n$, so $i \notin \limsup A_n,i \notin \liminf A_n$.
Let $x=\frac{p}{q}$. Then, $x (kq) = \frac{p(kq)}{q} = pk \in \mathbb{N}$, so $x \in A_{kq} \forall k \in \mathbb{N}$.
Now, given  $m \in \mathbb{N}$,there is some $k$ such that $kq > m$. So, for all $m$, $x \in \bigcup_{l \geq m} A_l$, taking $l=kq$. Since the above applies for all $m$, it follows that $x \in \displaystyle\bigcap_{m=1}^\infty \bigcup_{l \geq m} A_l = \limsup A_m$. Since $x$ was a rational number, and we already showed that irrationals can't be part of $A_n$, it follows that $\limsup A_n = \mathbb{Q}$.
The other way, suppose that $x = \frac{p}{q} \in \liminf A_n  =\displaystyle\bigcup_{n=1}^\infty \bigcap_{k \geq n} A_k$. This means that $x \in \bigcap_{k \geq n} A_k$ for some natural number $N$. This means, that $x(N+2)=\frac{p(N+2)}{q}$ and $x(N+1)=\frac{p(N+1)}{q}$ are both integers. By subtraction, we get that $\frac{p}{q}$ is also a natural number.  But then, $x$ itself is a natural number. So every number in the $\liminf$ is a natural number.
On the other hand, if we have a natural number $q$, then $q \in A_n$ for all $n$, because taking $m= qn$, we see that $q = \frac{qn}{n}$, so $q \in A_n$. Since this applies for all $n$, certainly $q \in \bigcap_{n \geq 1} A_n$, and hence $q \in \displaystyle\bigcup_{n=1}^\infty \bigcap_{k \geq n} A_k$. Thus, all natural numbers are in the $\liminf$.
Thus, $\liminf A_n= \mathbb{N}$. I hope this was a clearer solution than the previous one.
