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Math Help - Analysis subsequential limits

  1. #1
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    Analysis subsequential limits

    I need to prove

    i) lim n to infinity sup Sn is an element of SL(Sn)
    ii) same thing but replace sup with inf


    I know SL(Sn) is the set of all limits of all convergent subsequences of Sn.

    Also, lim n to infinity sup Sn can be defined as sup SL(Sn)

    So it may be easier to show that Sup SL(Sn) is an element of SL(Sn)

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  2. #2
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    Sorry, not a specialist in this area, but what is Sn? Is it just a sequence of real numbers S1, S2, ...? If so, then what is sup Sn?
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  3. #3
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    Can only help you if I can understand your question. Explain or state the definition of what Sn and SL(Sn) are.
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  4. #4
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    Sn is sequence of real numbers

    SL(Sn) is the set of all limits of all convergent subsequences of the sequence Sn
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  5. #5
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    Yes, but what is sup Sn? Is it a constant sequence where each element is equal to the supremum of the sequence \{S_n\}_{n=0}^\infty? Or is it a supremum of an individual number S_n? Or maybe, what I think is likely, it is \sup_{m\ge n}S_m, as in here, so the whole expression is \displaystyle\limsup_{n\to\infty}S_n := \lim_{n\to\infty}(\sup_{m\geq n}S_m)?
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  6. #6
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    Quote Originally Posted by emakarov View Post
    Yes, but what is sup Sn? Is it a constant sequence where each element is equal to the supremum of the sequence \{S_n\}_{n=0}^\infty? Or is it a supremum of an individual number S_n? Or maybe, what I think is likely, it is \sup_{m\ge n}S_m, as in here, so the whole expression is \displaystyle\limsup_{n\to\infty}S_n := \lim_{n\to\infty}(\sup_{m\geq n}S_m)?

    It is the last definition that you gave with the link
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  7. #7
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    Let's prove that there exists a subsequence S_{n_1}, S_{n_2},\dots whose limit is \lim_{n\to\infty}(\sup_{m\geq n}S_m). Let T_n=\sup_{m\geq n}S_m and let a=\lim_{n\to\infty}T_n. We can ensure that for every i and every j>i, |S_{n_j}-a|<1/i, i.e., the entire tail of the subsequence starting from S_{n_i} lies in the 1/i-neighborhood of a.

    Indeed, suppose i is given and S_{n_1}, \dots, S_{n_{i-1}} have already been chosen. From the definition of a, there is a point m such that

    for all n\ge m, |T_n-a|<1/(2i) (*)

    Without loss of generality, we can assume that m>n_{i-1}; otherwise, choose a new m as n_{i-1}+1; then (*) still holds. In particular, |T_{m}-a|<\1/(2i), i.e., |(\sup_{m\ge m}S_m)-a|<1/(2i). From the definition of sup, there exists an index, which we'll call n_i, such that n_i\ge m>n_{i-1} and |S_{n_i}-\sup_{m\ge m}S_m|<1/(2i). Thus, |S_{n_i}-a|<1/i.

    TL;DR: We choose a subsequence whose limit is the limit superior of \{S_n\} (call it a). Since a is a limit, sups of tails of \{S_n\} come close to a. Since those are sups, there are individual elements S_n that come close to a.
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