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limit of a subsequence

  1. Oct 11, 2009 #1
    1. The problem statement, all variables and given/known data
    Show that the lim sup of a bounded sequence is a limit of a subsequence.


    2. Relevant equations
    Sequence: Sn
    Subsequence: Snk

    3. The attempt at a solution
    An existent lim sup means that at a large enough N, the subsequence could hug the bottom of the lim sup to within epsilon (e). I don't know how to formalize this notion.

    The last two steps of the proof are likely
    -e < Snk - lim sup Sn < e
    |Snk - lim sup Sn| < e
     
  2. jcsd
  3. Oct 12, 2009 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    The lim sup is the sup of the set of all subsequential limits. That means that, calling the lim sup a, given any [itex]\epsilon> 0[/itex] there exist a subsequential limit within [itex]\epsilon[/itex] of a (and less than a). In particular, for every positive integer m, there exist a subsequential limit within 1/2m of a. And because that is a limit of a subsequence, there exist a member of that subsequence, call it [itex]a_{m}[/itex], within 1/2m of that subsequential limit. Look at what the subsequence [itex]a_m[/itex] converges to. You are forming a new subsequence by, essentially, taking a member from every subsequence.
     
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