Set lim inf and lim sup question. If $A_n = \{m/n : m \in N\}, n \in N$. What is $\lim \inf_n$ $A_n$ and $\lim\sup_n$ $A_n$?
Using the definitions:
$\lim \inf A_n = \bigcup\limits_{n\in N}\bigcap\limits_{k > n} m/n = \bigcup\limits_{n\in N}0 = 0$
$\lim \sup A_n =\bigcap\limits_{n\in N}\bigcup\limits_{k > n}
\frac{m}{n} = \bigcup\limits_{n\in N} \frac{m}{n}= 0$
I am trouble thinking of the intersections and unions. For instance $\bigcap\limits_{k > n}$ $\frac{m}{n}$ is the intersection of $\{\frac{m}{n},\frac{m}{n+1},\frac{m}{n+2},....\}$ I see how this eventually approaches $0$, but this is the intersection of a bunch of points instead of sets? Essentially the union and intersection operations don't make much sense to me in this context because I'm dealing with points instead of sets, but I don't know how else to approach the problem.
 A: lim sup$_n$ $A_n$ = $\mathbb{Q}$.To see let $x \in$ lim sup$_n$ $A_n$. Then $x = \frac{m_n}{n}$ for infinitely many $n$, so $x \in \mathbb{Q}$, thus lim sup$_n$ $A_n$ $\subset \mathbb{Q}$.
Now let $x \in \mathbb{Q}$. So $x = \frac{m}{n}$ for integers $m$ and $n \neq 0$. This means $x = \frac{cm}{cn}$ for $c \geq 1$, so $x \in$ lim sup$_n$ $A_n$. Thus $\mathbb{Q} \subset$ lim sup$_n$ $A_n$. So lim sup$_n$ $A_n$ = $\mathbb{Q}$.
lim inf$_n$ $A_n$ = $\mathbb{N}$. 
lim inf$_n$ $A_n$ = $\{xn \in \mathbb{N}\}$, so $\mathbb{N} \subset$ lim inf$_n$ $A_n$.
To see the reverse direction, we use the fact lim inf$_n$ $A_n$ $\subset$ lim sup$_n$ 
 $A_n$.
So for  $x \in$ lim inf$_n$ $A_n$ $\subset$ lim sup$_n$ 
 $A_n$ = $\mathbb{Q}$. Let $x = \frac{p}{q}$ for natural numbers $p$ and $q$. Since $x$ is in $A_n$ let's say for $n \geq n_0$, choose a $n_1 > n_0$, so $n_1 \frac{p}{q} \in \mathbb{N}$. Also, $(n_1 + 1)\frac{p}{q} = n_1 \frac{p}{q} + \frac{p}{q}\in \mathbb{N}$. Since $n_1 \frac{p}{q}\in \mathbb{N} $, $\frac{p}{q}$ must also be in $\mathbb{N}$ in order for their sum to be in $\mathbb{N}$. Thus $x = \frac{p}{q} \in \mathbb{N}$ completing the proof.
