Existence of limits of group elements

Question

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?

Answer 1

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:

1. $x \in A, x \in B$. 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 $x \notin A_n \Delta B_n$.

2. $x \in A, x \notin B$. 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 $x \in A_n \Delta B_n$.

3. $x \notin A, x \in B$. 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 $x \in A_n \Delta B_n$.

4. $x \notin A, x \notin B$. 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 $$\liminf A_n \Delta B_n =A \Delta B \\ \limsup A_n \Delta B_n =A \Delta B$$