Is it true that $\cup_{n=N}^{\infty}A_{n}\setminus\cap_{n=N}^{\infty}A_{n}=\cup_{n=N}^{\infty}A_{n}\triangle A_{n+1}$? During an exam I have claimed that if $\{A_{n}\}_{n=1}^{\infty}$
  then for any $N\in\mathbb{N}$
  $$\limsup A_{n}\setminus\liminf A_{n}\subseteq\cup_{n=N}^{\infty}A_{n}\setminus\cap_{n=N}^{\infty}A_{n}=\cup_{n=N}^{\infty}A_{n}\triangle A_{n+1}$$
The equality part of my statement was marked as wrong, but I am having difficulty to understand if it is indeed wrong, and if I can replace $=$
  with $\subseteq$
  to make it correct.
My argument for the first the first containment is the description of $\limsup A_{n},\liminf A_{n}$
  as the set of all elements of in those sets that are in infinity many sets and as those that are not in a finite number of those $A_{n}$
 .
For the second part, I tried to understand what is $$(A\triangle B)\cup(B\triangle C)$$
  and using a diagram I got that it is $$(A\cup B\cup C)\setminus(A\cap B\cap C)$$
  so I figured I can deduce the equality for the infinite case as well, although I can't prove it formally. 
Is it true that $$\limsup A_{n}\setminus\liminf A_{n}\subseteq\cup_{n=N}^{\infty}A_{n}\setminus\cap_{n=N}^{\infty}A_{n}=\cup_{n=N}^{\infty}A_{n}\triangle A_{n+1} ?$$
  or at least that $$\limsup A_{n}\setminus\liminf A_{n}\subseteq\cup_{n=N}^{\infty}A_{n}\setminus\cap_{n=N}^{\infty}A_{n}\subseteq\cup_{n=N}^{\infty}A_{n}\triangle A_{n+1} ?$$
Any help is greatly appreciated!
 A: You are asking about the relation between sets $L=\bigcup\limits_{n=N}^{\infty}A_{n}\setminus\bigcap\limits_{n=N}^{\infty}A_{n}$ and $R=\bigcup\limits_{n=N}^{\infty}A_{n}\triangle A_{n+1}$. (The letters $L$ and $R$ stand for LHS and RHS of your in your proposed equality.)

$\boxed{L\subseteq R}$
If $x\in L$, this definitely implies that $x\in A_n$ for some $n\ge N$. Let $n_0$ be the smallest integer with the properties $x\in A_{n_0}$ and $n_0\ge N$. 
If $n_0>N$ then $x\notin A_{n_0-1}$ and thus $x\in A_{n_0}\triangle A_{n_0-1}$, which means that $x\in R$.
The other possibility is that $n_0=N$. There exists at least one $n$ such that $x\notin A_n$ (since $x\notin \bigcap\limits_{n=N}^{\infty}A_{n}$). Now we denote by $n_1$ the smallest integer such that $x\notin A_{n_1}$ and $n_1\ge N$. (We already know that, in this case, $n_1>n_0=N$.)  So we have $x\in A_{n_1}\triangle A_{n_1-1}$ and, again, $x\in R$.

$\boxed{R\subseteq L}$
If $x\in R$ then $x\in A_n\triangle A_{n+1}=(A_n\setminus A_{n+1})\cup(A_{n+1}\setminus A_n)$ for some $n\ge N$.
If we assume that $x\in A_n\setminus A_{n+1}$ then $x\in\bigcup\limits_{n=N}^{\infty}A_{n}$ (since $x\in A_n$) and $x\notin \bigcap\limits_{n=N}^{\infty}A_{n}$ (since $x\notin A_{n+1}$).
The case $x\in A_{n+1}\setminus A_n$ is similar.

About limit superior and limit inferior:
If I remember the the definition of $\liminf$ and $\limsup$ correctly, we have
$$\limsup A_n = \bigcap\limits_{N=1}^\infty \bigcup\limits_{n=N}^\infty A_n\\
\liminf A_n = \bigcup\limits_{N=1}^\infty \bigcap\limits_{n=N}^\infty A_n$$
So for any fixed $N_0$ you have
$$\limsup A_n \setminus \liminf A_n =
\left(\bigcap\limits_{N=1}^\infty \bigcup\limits_{n=N}^\infty A_n\right)\setminus\left(\bigcup\limits_{N=1}^\infty \bigcap\limits_{n=N}^\infty A_n\right)\subseteq 
\bigcup\limits_{n=N_0}^\infty A_n\setminus\left(\bigcup\limits_{N=1}^\infty \bigcap\limits_{n=N}^\infty A_n\right)\subseteq 
\bigcup\limits_{n=N_0}^\infty A_n \setminus \bigcap\limits_{n=N_0}^\infty A_n.$$
