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Lim inf and lim sup

  1. Feb 10, 2005 #1
    So if the sequence Rn is the list of all rational numbers in (0,1), how am I to prove that lim inf=0 and lim sup= 1?


    I understand that this sequence has sup=1 and inf=0, but since I do not know what the sequence will look like as n approaches infinity, how do I know lim sup=1 and lim inf=1?

    For example, if the sequence is non-decreasing, then can I necessarily say that lim inf=0. I know that the inf will be 0, but lim inf cannot be approaching this number since the rational numbers are increasing. Tell me where my thinking is wrong.
     
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  3. Feb 10, 2005 #2

    mathwonk

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    it helps if you have a definition of limsup and liminf.

    i do not know the usual definition, but i would guess that you could say the lim sup is the smallest number L such that for all e>0, only a finite number of elements of the sequence lie above L+e.

    then it seems clear that your problem is true.
     
  4. Feb 11, 2005 #3

    Galileo

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    The sequence of infima must be nondecreasing and the sequence of suprema must be nonincreasing.
    They are clearly both bounded (below by 0 and above by 1), therefore they are convergent.

    Easiest way to show the lim inf is 1 may be to assume it is less than 1 a derive a contradiction and to assume it is greater that 1 and derive a contradiction. (same for supremum).

    EDIT: Eh, I think I`m confusing stuff. I have to check my definition of lim inf, because lim inf should be 0, not 1...
     
    Last edited: Feb 11, 2005
  5. Feb 11, 2005 #4

    HallsofIvy

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    "lim inf" of a sequence is the infimum of the set of all subsequence limits of the sequence.

    "lim sup" of a sequence is the supremum of the set of all subsequence limits of the sequence.

    To show that the lim sup of the set (sequence since it is countable) of all rational numbers between 0 and 1 is 1 you must show that there exist some sequence of rational number in that set that converges to 1.

    To show that the lim inf of the set (sequence since it is countable) of all rational numbers between 0 and 1 is 0 you must show that there exist some sequence of rational number in that set that converges to 0.

    Hint: Any real number is the limit of some sequence of rational numbers.
     
  6. Feb 11, 2005 #5

    mathwonk

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    ok, if the lim sup is the sup of the set of all subsequence limits, and if only a finite number of elements of the sequence lie above L+e for every e>0, then certainly no number larger than L+e is the limit of a subsequence. Hence my L is ≥ the lim sup.

    But if we choose any e > 0 and consider L-e, then by my definition of L, and since L-e is smaller than L, certainly an infinite number of elements of the sequence are greater than L-e. Hence there is an infinite number of elements of the sequence caught between L-e and L, for every e>0.

    Then it is obvious that L is a subsequence limit. So L is actually the largest subsequence limit. Hence L is the limsup.


    Now since for every e>0, only a finite number of elements of the sequence of all rationals in (0,1) lies above L+e, we must have L ≥ 1. And since L is a limit of rational between 0 and 1, also L ≤ 1. so L = 1.

    My definition seems so much easier than the official one. What gives?

    Maybe they differ in the case where the limsup is minus infinity or something, which is irrelevant here.
     
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