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Math Help - liminf to limsup transformation question..

  1. #1
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    liminf to limsup transformation question..

    y_n is a bounded sequence

    liminf(-y_n)=-limsup(y_n)
    n->infinity

    i cant understand how it happens??
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  2. #2
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    Quote Originally Posted by transgalactic View Post
    y_n is a bounded sequence

    liminf(-y_n)=-limsup(y_n)
    n->infinity

    i cant understand how it happens??
    If A be a non-empty bounded subset of \mathbb{R} then \sup (-A) = -\inf A \text{ and }\inf (-A) = -\sup A.
    This is the key fact that we need. (Notation: -A = \{ - a|a\in A\})

    Let r = \liminf (-y_n).
    Therefore, r = \sup \left\{ \inf\{ -y_k |k\geq n\} | k\geq 1\right\}
    But then, -r = \inf \left\{ \boxed{- \inf \{ -y_k |k\geq n\}} | k\geq 1\right\}
    Apply again to the boxed expression, -r = \inf \left\{ \sup \{ y_k |k\geq n\} | k\geq 1 \right\}
    Thus, -r = \limsup (y_n).
    Putting this together we get, \liminf (-y_n) = - \limsup (y_n).
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    You may want to wait for confirmation on this...it may not be rigorous enough....also how late it is may cause concern for incorrectness.

    I think you alternatively to TPH's response may be able to say that since x_n is a bounded sequence that the set of subsequential limits S of x_n is a bounded subset of the reals. This fact lets us apply all the basic "rules and theorems" of subsets of the reals to S. So now using TPH's first remark and noting that \limsup x_n=\sup S and \liminf x_n=\inf S we can draw our conclusion.
    Last edited by Mathstud28; January 20th 2009 at 12:57 AM. Reason: Ambiguous wording
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