Results 1 to 3 of 3

Math Help - Limit of sequence of sets

  1. #1
    Newbie
    Joined
    Aug 2012
    From
    north carolina
    Posts
    8

    Limit of sequence of sets

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Sep 2012
    From
    Washington DC USA
    Posts
    525
    Thanks
    147

    Re: Limit of sequence of sets

    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.
    Last edited by johnsomeone; September 18th 2012 at 03:10 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2012
    From
    north carolina
    Posts
    8

    Re: Limit of sequence of sets

    Quote Originally Posted by johnsomeone View Post
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Strongly decreasing sequence of sets
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: November 14th 2010, 05:50 PM
  2. Replies: 2
    Last Post: October 26th 2010, 11:23 AM
  3. Sequence of leesgue measurable sets
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: April 10th 2010, 10:15 AM
  4. Proof convergence sequence of sets
    Posted in the Discrete Math Forum
    Replies: 8
    Last Post: February 26th 2010, 10:06 AM
  5. Replies: 1
    Last Post: October 18th 2008, 12:56 PM

Search Tags


/mathhelpforum @mathhelpforum