Proving $\limsup (A_n) \cap \limsup (A^{c}_{n}) = \limsup (A_n \cap A^{c}_{n+1})$ 
Let $(\Omega, \mathscr{F},P)$ be a probability space and $\{A_n\}_{n\geq 1}$ a sequence in $\mathscr{F}$. Prove that $$\limsup (A_n) \cap \limsup (A^{c}_{n}) = \limsup (A_n \cap A^{c}_{n+1})$$

Since is an equality I have to show both inclusions. 
Suppose  $x\in \limsup (A_n) \cap \limsup (A^{c}_{n})$.
According to the definition of limsup:
$x \in \limsup (A_n)$ $\dashrightarrow$ $\forall n \ge 1,  \ \exists k \ge n$ such that $ x \in A_{k}$ 
$x \in \limsup (A^{c}_n)$ $\dashrightarrow$ $\forall m \ge 1,  \ \exists h \ge m$ such that $ x \in A^{c}_{h}$
From here I don't know how to procede. 
 A: Let me give a proof using events.
Borrow the notion "infinitely often" (i.o.) from Durrett's probability textbook (in the section on Borel-Cantelli's lemma), rewrite $\limsup$ as $i.o.$, and the reverse inclusion should be obvious.
$$\{A_n \cap A_{n+1}^c i.o.\} \subseteq \{A_n i.o.\}  \cap  \{A_n^c i.o.\}$$
Let $x\in\{A_n \cap A_{n+1}^c i.o.\}$ for any $n$, find $N\ge n$ so that $x\in A_N \cap A_{N+1}^c$.  This shows the $x \in \{A_n i.o.\}$.  Similarly, $x \in \{A_n^c i.o.\}$

Show the direct inclusion by contradiction.  Let $x \notin \{A_n \cap A_{n+1}^c i.o.\}$.  i.e.  $x \in A_n \cap A_{n+1}^c$ "finitely often".
Therefore, $x \in A_n^c \cup A_{n+1}$ for sufficiently large $n$.  Since $x \in \{A_n i.o. \}$, there exists a (larger) $N\ge n$ so that $x \in A_N$.  Invoke our assumption $x \in A_N^c \cup A_{N+1}$ to conclude that $ x\in A_{N+1}$.  Inductively, we've shown that $x \in A_m$ for all $m \ge N$, which contradicts $x \in \{A_n^c i.o.\}$.
A: We can use the set-theoretic formulation: $\limsup A_n=\left ( \bigcap_{n\ge 1}\bigcup_{j\ge n}A_j \right ).$ That is to say, $x\in \limsup A_n$ if and only if for all integers $n$ there is an integer $j\ge n$ such that $x\in A_j.$ 
So, suppose $x\in \limsup A_n\cap \limsup A_{n}^c.$ Fix $n$ and choose the least integer $j\ge n$ for which $x\in A^c_{j+1}.$ Then, $x\notin A_j^c\Rightarrow x\in A_j\Rightarrow x\in A_j\cap A^c_{j+1}$ and it follows that $x\in \limsup (A_n\cap A^c_{n+1}).$
On the other hand, if $x\in \limsup (A_n\cap A^c_{n+1})$, then for each integer $n$, there is an integer $j\ge n$ such that $x\in A_j\cap A^c_{j+1}$ and it follows that $x\in \limsup A_n\cap \limsup A_{n}^c$.
