Prove $\limsup A_n \smallsetminus \liminf A_n = \limsup (A_n \cap A_{n+1})$ We have a sequence of sets indexed on the naturals and we want to show the following is true.  There is no assumption of increasing or decreasing containment so I'm not sure what you're supposed to do with the consecutive-ness.
$\limsup A_n \smallsetminus \liminf A_n = \limsup (A_n \cap A_{n+1})$
Note that minus is set theoretic difference.
An obvious issues is that the LHS will not separate nicely since Limsup doesn't yield strict equality when broken over intersection...I tried pushing around DeMorgan's Law a bit but I'm stumped.
 A: The stated equality does in general not hold, as illustrated by the example
$$A_n = \begin{cases}[0,1] &, n \equiv 0 \pmod{2}\\ [2,3] &, n \equiv 1 \pmod{2}.\end{cases}$$
We have $\limsup A_n = [0,1] \cup [2,3]$ and $\liminf A_n = \varnothing = \limsup (A_n \cap A_{n+1})$ then.
However, we always have $\DeclareMathOperator{\sdiff}{\triangle}$
$$\limsup A_n \setminus \liminf A_n = \limsup (A_n \sdiff A_{n+1})$$
where $A \sdiff B$ is the symmetric difference $(A\setminus B) \cup (B\setminus A)$.
A: You know that $x\in\limsup_{n\in\Bbb N}A_n$ iff $\{n\in\Bbb N:x\in A_n\}$ is infinite, and $x\in\liminf_{n\in\Bbb N}A_n$ iff $\{n\in\Bbb N:x\notin A_n\}$ is finite. Thus,
$$x\in\limsup_{n\in\Bbb N}A_n\setminus\liminf_{n\in\Bbb N}A_n$$
iff $\{n\in\Bbb N:x\in A_n\}$ and $\{n\in\Bbb N:x\notin A_n\}$ are both infinite.
On the other hand, $x\in\limsup_{n\in\Bbb N}(A_n\cap A_{n+1})$ iff $\{n\in\Bbb N:x\in A_n\cap A_{n+1}\}$ is infinite. Suppose that we can arrange matters so that there is an $x$ such that $x\in A_n$ iff $n$ is even; then for all $n\in\Bbb N$ we’ll have $x\notin A_n\cap A_{n+1}$ and therefore $x\notin\limsup_{n\in\Bbb N}(A_n\cap A_{n+1})$, but we clearly will have $x\in\limsup_{n\in\Bbb N}A_n\setminus\liminf_{n\in\Bbb N}A_n$. And this is easy to arrange: for example, just let
$$A_n=\begin{cases}
\{0\},&\text{if }n\text{ is even}\\
\varnothing,&\text{if }n\text{ is odd}\;.
\end{cases}$$
To find a correct equality, suppose that $x\in\limsup_{n\in\Bbb N}A_n\setminus\liminf_{n\in\Bbb N}A_n$, so that $\{n\in\Bbb N:x\in A_n\}$ and $\{n\in\Bbb N:x\notin A_n\}$ are both infinite, and let $$M=\{n\in\Bbb N:x\in A_n\setminus A_{n+1}\}\;;$$ then $M$ must be infinite (why?), and we must have $x\in\limsup_{n\in\Bbb N}(A_n\setminus A_{n+1})$. Thus,
$$\limsup_{n\in\Bbb N}A_n\setminus\liminf_{n\in\Bbb N}A_n\subseteq\limsup_{n\in\Bbb N}(A_n\setminus A_{n+1})\;.$$
The reverse inclusion is even easier to check, so
$$\limsup_{n\in\Bbb N}A_n\setminus\liminf_{n\in\Bbb N}A_n=\limsup_{n\in\Bbb N}(A_n\setminus A_{n+1})\;.$$
