Limit of sequence of sets

**sepmo** · August 24th 2012, 04:12 PM

I am trying to prove a proposition, but it's proving harder than I expected. I was wondering if someone could lead me in the right direction. Please don't give full answers. I'm just looking for a hint. The problem says:
If $\{E_n\}$ is a sequence of sets and $D_1=E_1$, $D_{n+1}=D_n\bigtriangleup E_{n+1}, n=1,2,...$ , show that $lim D_n$ existis if and only if $lim E_n = \emptyset$ .

Thank you.

**johnsomeone** · September 18th 2012, 01:36 PM

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.
(And it appears my suspicion proved wrong. Typical.

)

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

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

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

Proof: Will be by induction.

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

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

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

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

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

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

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

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

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

But those three statements ( $S(n)$ true, $x \in D_{2n+1}$ , and $x \notin D_{2n+2}$ ) together exactly comprise the statement $S(n+1)$ .

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

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

Claim #2: If $x \in \mathbb{X}$ and there exists $N \in \mathbb{N}$ such that $x \notin E_n \ \forall \ n > N$ (always here $n \in \mathbb{N}$ ),

then $x \in D_N \Rightarrow x \in D_n \ \forall \ n \ge N$ ,

and $x \notin D_N \Rightarrow x \notin D_n \ \forall \ n \ge N$ .

Proof:

If $x \in D_N$ , then since $x \notin E_{N+1}$ , have that $x \in D_{N+1}=D_N\bigtriangleup E_{N+1}$ .

Next, since $x \in D_{N+1}$ and $x \notin E_{N+2}$ , can conclude that $x \in D_{N+2}=D_{N+1}\bigtriangleup E_{N+2}$ .

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

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

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

That proves Claim #2.

**sepmo** · September 18th 2012, 01:45 PM

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

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

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

Proof: Will be by induction.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Proof:

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

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

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

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

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

That proves Claim #2.

Thanks for the response. I actually found a way of proving it using the characteristic function. It is quite a short proof.

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