Existence of limits of group elements Referring to the usual abelian group ($2^X, \Delta$) defined as the power set of a set with the symmetric difference as the multiplication operation. Let ($A_n \in 2^X)_{n \in \mathbb{N}}$ be a sequence of sets. 
$\lim A_n$ exists iff 
lim sup $A_n$ = lim inf $A_n = A$. 
If lim $A_n$ and lim $B_n$ exist, what can I say about lim $A_n \Delta B_n$ and on what basis?    
 A: Note first that $x \in \limsup A_n$ if and only if, for each $N$, there exists some $n >N$ so that $x \in A_n$. 
Same way, $x \in \liminf A_n$ if and only if, there exists some $N$, such that, for all $n >N$ we have $x \in A_n$. 
Now, since $\lim A_n$ exists, the above implies the following:
For each $x \in X$ then exactly one of the following hold:


*

*$x \in \lim_n A_n$. Then, there exists some $N$, such that, for all $n >N$ we have $x \in A_n$. This comes from $x \in \liminf A_n$.

*$x \notin \lim_n A_n$. Then, there exists some $N$, such that, for all $n >N$ we have $x \notin A_n$. This comes from $x \notin \limsup A_n$.


Now, your problem
Denote $A=\lim A_n, B= \lim B_n$. You have 4 possible scenarios:


*

*$x \in A, x \in B$. Therefore, there exists some $N_1, N_2$ such that $x \in A_n \forall n >N_1$ and $x \in B_n$ forall $n >N_2$.


Then, for all $n > \max\{ N_1, N_2 \}$ you have $x \notin A_n \Delta B_n$.


*$x \in A, x \notin B$. Therefore, there exists some $N_1, N_2$ such that $x \in A_n \forall n >N_1$ and $x \notin B_n$ forall $n >N_2$.


Then, for all $n > \max\{ N_1, N_2 \}$ you have $x \in A_n \Delta B_n$.


*$x \notin A, x \in B$. Therefore, there exists some $N_1, N_2$ such that $x \notin A_n \forall n >N_1$ and $x \in B_n$ forall $n >N_2$.


Then, for all $n > \max\{ N_1, N_2 \}$ you have $x \in A_n \Delta B_n$.


*$x \notin A, x \notin B$. Therefore, there exists some $N_1, N_2$ such that $x \notin A_n \forall n >N_1$ and $x \notin B_n$ forall $n >N_2$.


Then, for all $n > \max\{ N_1, N_2 \}$ you have $x \notin A_n \Delta B_n$.
From here, it is trivial to deduce that 
$$\liminf A_n \Delta B_n =A \Delta B \\
\limsup A_n \Delta B_n =A \Delta B$$
