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Lim sup, lim inf definition/convention

  1. May 11, 2007 #1

    quasar987

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    My book says that if the set of all cluster points is empty, then we write lim sup = [itex]-\infty[/itex] and if the sequence is not bounded above, we write limsup = [itex]+\infty[/itex].


    But what if both happen at the same time? for instance consider x_n=1/n. There are no accumulation points and it is unbounded above.
     
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  3. May 11, 2007 #2

    NateTG

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    Maybe I'm getting too tired, but
    [tex]1>\frac{1}{n}[/tex]
    is an upper bound, and
    [tex]0[/tex]
    is an accumulation point.

    You might consider [itex]x_n=n(-1)^n[/itex] which is unbounded in the reals, and doesn't have any real accumulation points. However, in the extedended reals, [itex]\pm \infty[/itex] are cluster points of that sequence, and [itex]+ \infty[/itex] is an upper bound for all sequences.
     
  4. May 11, 2007 #3

    quasar987

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    It is I who is too tired. I meant to say x_n = n.
     
  5. May 11, 2007 #4

    HallsofIvy

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    I don't see a problem. Since the set of all cluster points is empty, lim sup= [itex]-\infty[/itex] and since the sequence is not bounded above, limsup= [itex]\infty[/itex].
     
  6. May 11, 2007 #5

    quasar987

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    Well, actually, the author give the same x_n=n as an example and he write lim sup=+[itex]\infty[/itex].

    As if the fact that it is not bounded above takes priority over the fact that the set of all cluster points is empty.
     
  7. May 11, 2007 #6

    AKG

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    Are "lim sup" and "limsup" supposed to be two different things?
     
  8. May 11, 2007 #7

    quasar987

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    No, no.

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