Show that $s\in\limsup\limits_{n\to\infty} A_n$ iff there are infinitely many $n\in\mathbb{N}$ such that $s\in A_n$. Let $(S, \mathcal{A}, \mu)$ be a measure space. For $A_1,A_2,\ldots\subseteq S$ define $$\limsup_{n\to\infty} A_n = \bigcap_{k=1}^\infty\bigcup_{n= k}^\infty A_n$$
Exercise: Show that $s\in\limsup\limits_{n\to\infty} A_n$ if and only if there are infinitely many $n\in\mathbb{N}$ such that $s\in A_n$.
What I've tried:
I think I have prove the statement in the $(\Rightarrow)$ direction. Let $s\in\limsup\limits_{n\to\infty}A_n$. Suppose that there are only a finite number of $n$ for which $s\in A_n$, and let $N$ denote the largest $n\in\mathbb{N}$ such that $s\in A_N$. We have that $$\limsup\limits_{n\to\infty} A_n \subseteq (A_1\cup A_2\cup\ldots)\cap (A_2\cup A_3\cup\ldots)\cap\ldots \cap (A_N\cup A_{N+1}\cup\ldots)\cap(A_{N+1}\cup A_{N+2}\cup\ldots)\subseteq (A_{N+1}\cup A_{N+2}\cup\ldots)$$ Now, since $s\not\in (A_{N+1}\cup A_{N+2}\cup\ldots)$ we have that $s\not\in\limsup\limits_{n\to\infty}A_n$. This is a contradiction, so we must have that there are infinitely many $n\in\mathbb{N}$ such that $s\in A_n$.
I have thought about how I should prove the $(\Leftarrow)$ direction, but my results thus far aren't very rigorous. Suppose that there are infinitely many $n\in\mathbb{N}$ such that $s\in A_n$. Then for every $k$ we have that $s\in\bigcup\limits_{n=k}^\infty A_n$, and since $\limsup\limits_{n\to\infty}A_n$ is the intersection of those $\bigcup\limits_{n=k}^\infty A_n$, we have that $s\in\limsup\limits_{n\to\infty}A_n$. 
Question: How should I show this exercise?
Thanks!
 A: Right to left direction. Given such an $x$ and $k$, we are to show that $x\in\displaystyle\bigcup_{n\geq k}A_{n}$.
Since $x\in A_{i}$ for infinitely many $i$, the set $I:=\{i: x\in A_{i}\}$ is infinite. Choose an $i_{0}\in I$, then $I-\{1,...,i_{0}\}\ne\emptyset$, if not, then $I$ is a finite set. So pick an $i_{1}\in I-\{1,...,i_{0}\}$, then $I-\{1,...,i_{0},...,i_{1}\}\ne\emptyset$, so pick an $i_{2}\in I-\{1,...,i_{0},...,i_{1}\}$. Continue in this fashion, a strictly increasing sequence $\{i_{l}\}$ is such that $x\in A_{i_{l}}$. Since $\{i_{l}\}$ is a sequence of natural numbers, if it converges, then it is eventually constant, so it is not convergent, and hence $i_{l}\rightarrow\infty$, so some $l_{0}$ is such that $i_{l}\geq k$ for all $l\geq l_{0}$, but we have $x\in A_{i_{l}}$, so $x\in\displaystyle\bigcup_{n\geq k}A_{n}$.
A: OK. We should really try interpreting the meaning of all these set operations!
$\left ( \Rightarrow  \right )$: If $s\in\limsup\limits_{n\to\infty} A_n$, then what it means is that $s$ is in every $\bigcup_{n= k}^\infty A_n$. If the statement is not true, i.e. $s$ is only in finitely many $A_n$, there must exists a largest $n$ s.t. $s$ no longer belongs to $A_{n+1}$ and so on. Now what can you say?
$\left ( \Leftarrow  \right )$: Conversely, if $s$ is in infinitely many $A_n$'s, then it means $s$ is also in any `tail', i.e. the infinite union!!! Hence it is in their intersection!
Now, some bonus: if you really want to understand $\limsup$ and $\liminf$ for sets, try this example: Let $A_n = \left \{ \left ( -1 \right )^n \right \}$. Then $\limsup A_n = \left \{ \pm 1\right \}$, and $\liminf A_n = \varnothing $. Reinterpret: $\pm 1 \in \limsup A_n$ because both $\pm 1$ happen infinitely may times; yet none of them are in $\liminf A_n$ because they missed some $A_n$ infinitely many times!!!
A: Why by contradiction ?
\begin{align*}
s\in \limsup_{n\to \infty }A_n&\iff s\in \bigcap_{n\in\mathbb N}\bigcup_{m\geq n}A_m\\
&\iff \forall n\in\mathbb N, s\in \bigcup_{m\geq n}A_m\\
&\iff \forall n\in\mathbb N, \exists m_n\geq n: s\in A_{m_n}\\
&\iff s \in A_{n}\text{ for an infinite number of $n$}.
\end{align*}
Observation / explanations : $(m_n)_{n\in\mathbb N}$ is a sequence of natural numbers s.t. $$\lim_{n\to \infty }m_n=+\infty .$$
By construction, you also have that $$s\in \limsup_{n\to \infty }A_n\iff s\in \bigcap_{n\in\mathbb N}A_{m_n}.$$
The claim follow by the fact that $$\#\{m_n\mid n\in\mathbb N\}=\infty .$$
