Originally Posted by

**johnsomeone** I don't know exactly what "the limit of a sequence of sets" has been defined to mean in your problem. However, I can show you something that I think might prove this claim for your whatever your definition there is. I'll just state what I can prove, and then leave it to you to see if that, when considered in light of whatever definition you have of "convergence sequences of sets", proves this claim. I suspect this will do the job.

Given a set $\displaystyle \mathbb{X}$ and $\displaystyle \mathcal{E} = \{E_n\}_1^{\infty} \subset \mathcal{P}(\mathbb{X})$, define $\displaystyle D_1=E_1$, and then inductively $\displaystyle D_{n+1}=D_n\bigtriangleup E_{n+1}, n=1,2,...$.

Let $\displaystyle E^* = \bigcap_{n = 1}^{\infty}E_n$.

**Claim #1:** If $\displaystyle E^* \ne \emptyset$ and $\displaystyle x \in E^*$, then $\displaystyle x \in D_n \Leftrightarrow n$ is odd.

**Proof:** Will be by induction.

Note that $\displaystyle x \in E^* \Rightarrow x \in E_n \ \forall n \in \mathbb{N}$.

Now let $\displaystyle S(m)$ be the proposition "$\displaystyle k \in \mathbb{N} \ni 1 \le k \le 2m \Rightarrow (x \in D_k \Leftrightarrow k$ is odd)".

Since $\displaystyle x \in E_1$, have that $\displaystyle x \in D_1 = E_1$. But also $\displaystyle x \in E_2$, so it follows that $\displaystyle x \notin D_2 = D_1 \bigtriangleup E_2$.

Thus have proven that $\displaystyle x \in D_1$, and $\displaystyle x \notin D_2$. That proves that $\displaystyle S(1)$ is true.

__Assume__ $\displaystyle S(n)$ is true for some $\displaystyle n \in \mathbb{N}, n \ge 1$.

Let $\displaystyle k=2n$. Then $\displaystyle k$ is even, and $\displaystyle k \le 2n$. Thus $\displaystyle S(n)$ true implies that $\displaystyle x \notin D_{k} ( = D_{2n})$.

So $\displaystyle x \notin D_{2n}$, but also $\displaystyle x \in E_{2n+1}$. Thus $\displaystyle x \in D_{2n+1}=D_{2n} \bigtriangleup E_{n+1}$.

Now $\displaystyle x \in D_{2n+1}$, but also $\displaystyle x \in E_{2n+2}$. Thus $\displaystyle x \notin D_{2n+2}=D_{2n+1} \bigtriangleup E_{n+1}$.

Have shown that $\displaystyle S(n)$ true implies that $\displaystyle x \in D_{2n+1}$ and also that $\displaystyle x \notin D_{2n+2}$.

But theose three statements together exactly comprise the statement $\displaystyle S(n+1)$.

Thus $\displaystyle S(n) \Rightarrow S(n+1) \ \forall n \in \mathbb{N}$. Also have shown that $\displaystyle S(1)$ is true.

Thus by induction, have proven $\displaystyle S(n)$ is true $\displaystyle \forall n \in \mathbb{N}$. That proves the claim #1.

**Claim #2:** If $\displaystyle x \in \mathbb{X} \ni \exists N \in \mathbb{N} \ni \ \forall n \in \mathbb{N}, n > N, x \notin E_n$,

then $\displaystyle x \in D_N \Rightarrow x \in D_n \forall n \in \mathbb{N}, n > N$, and $\displaystyle x \notin D_N \Rightarrow x \notin D_n \forall n \in \mathbb{N}, n > N.$.

**Proof:**

If $\displaystyle x \in D_N$, then $\displaystyle x \notin E_{N+1} \Rightarrow x \in D_{N+1}=D_N\bigtriangleup E_{N+1}$.

Then $\displaystyle x \in D_{N+1}$ and $\displaystyle x \notin E_{N+1}$ again together imply that $\displaystyle x \in D_{N+2}=D_{N+1}\bigtriangleup E_{N+2}$.

By the obvious induction, $\displaystyle x \in D_n \ \forall n > N$.

Conversely, if $\displaystyle x \notin D_N$, then $\displaystyle x \notin E_{N+1} \Rightarrow x \notin D_{N+1}=D_N\bigtriangleup E_{N+1}$.

The induction is clear: for $\displaystyle n \ge N, x$ will be in neither $\displaystyle E_{n+1}$ (by the definition of $\displaystyle N$), nor $\displaystyle D_n$ (by induction), and therefore $\displaystyle x$ won't be in $\displaystyle D_{n+1}$.

That proves Claim #2.