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Math Help - lim sup and lim inf of a countable collection of sets

  1. #1
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    lim sup and lim inf of a countable collection of sets

    Hi all,

    Problem statement:
    To build a case of a countable collection of sets of real numbers for which the lim inf and lim sup are not equal.

    My Attempt:
    (I normally associate lim sup and lim inf with sequences so I'm finding it a bit hard to think in terms of sets (intervals))

    Anyway, let me call my countable collection An.
    Suppose that I have two countable collection of sets, Bm and Cm where:

    - Bm is a countable collection of non-empty decreasing closed sets (intervals) of reals with
    B1 = [x, y-e]
    - Cm is a countable collection of non-empty decreasing closed sets (intervals) of reals with C1 = [y+e, z]
    (with x < y < z; e > 0)

    And Ai = Bj if i is even and Ai = Cj if i is odd.

    As n goes to infinity, lim inf An = L1 is going to lie somewhere between x and y-e
    and lim sup An = L2 is going to lie somewhere between y+e and z
    So, L1 cannot be equal to L2.

    Is my example valid?
    Last edited by director; September 29th 2013 at 08:19 AM.
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  2. #2
    Super Member Rebesques's Avatar
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    Re: lim sup and lim inf of a countable collection of sets

    Quote Originally Posted by director View Post
    Hi all,

    Problem statement:
    To build a case of a countable collection of sets of real numbers for which the lim inf and lim sup are not equal.

    My Attempt:
    (I normally associate lim sup and lim inf with sequences so I'm finding it a bit hard to think in terms of sets (intervals))

    Anyway, let me call my countable collection An.
    Suppose that I have two countable collection of sets, Bm and Cm where:

    - Bm is a countable collection of non-empty decreasing closed sets (intervals) of reals with
    B1 = [x, y-e]
    - Cm is a countable collection of non-empty decreasing closed sets (intervals) of reals with C1 = [y+e, z]
    (with x < y < z; e > 0)

    And Ai = Bj if i is even and Ai = Cj if i is odd.

    As n goes to infinity, lim inf An = L1 is going to lie somewhere between x and y-e
    and lim sup An = L2 is going to lie somewhere between y+e and z
    So, L1 cannot be equal to L2.

    Is my example valid?



    I don't think this example is right.
    Try something like this.
    Let the countable family be A_{m,n}=\left(\frac{2}{m+n},\frac{3}{m}\right), m,n naturals.

    Then \limsup A_{m,n} = \max A_{m,n}= A_{1,1}=(1,3) but \liminf A_{m,n} =\lim_{n\rightarrow\infty,m\rightarrow\infty}A_{m,  n}=\{0\}.
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  3. #3
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    Re: lim sup and lim inf of a countable collection of sets

    the definitions of liminf/limsup for a sequence of set is not exactly the same you are used for a sequence of natural numbers.

    \liminf_{n\to\infty} A_n = \bigcup_{N=1}^\infty \bigcap_{n\ge N} A_n and \limsup_{n\to\infty} A_n = \bigcap_{N=1}^\infty \bigcup_{n\ge N} A_n

    If you think about this in these terms the most simple exemple of a countable set with different limits should be An = [-1/n, 1,n], where limsup is (-1,1), liminf is {0}
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  4. #4
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    Re: lim sup and lim inf of a countable collection of sets

    A point x is in \liminf_{n\to \infty}A_n if there exists N \in \mathbb{N} such that for all n\ge N, x \in A_n. That same point x will be in \limsup_{n\to \infty}A_n if for all N\in \mathbb{N}, there exists n\ge N such that x \in A_n.

    In the OP example, if B_m = [x,y-e/m] and C_m = [y+e/m,z], then \liminf_{n\to \infty} A_n = \emptyset, \limsup_{n\to \infty}A_n = [x,z]\setminus {y}, so that could be what you are looking for.
    Last edited by SlipEternal; October 26th 2013 at 05:46 PM.
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