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Thread: subsequence definition question..

  1. #1
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    subsequence definition question..

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    Let $\displaystyle X_n = x_{r_n}$.
    Now, $\displaystyle \liminf x_n \leq \limsup x_n \leq \limsup X_n = \lim X_n$

    The reason why $\displaystyle \limsup X_n = \lim X_n$ is because $\displaystyle \{X_n\}$ is convergent.

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  3. #3
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    $\displaystyle
    \limsup X_n = \lim X_n
    $

    this expression is not correct
    X_n is bounded
    the limit of the sequence doesnt equal the limsup

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  4. #4
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    Quote Originally Posted by transgalactic View Post
    $\displaystyle
    \limsup X_n = \lim X_n
    $

    this expression is not correct
    X_n is bounded
    the limit of the sequence doesnt equal the limsup

    The sequence $\displaystyle X_n$ is convergent.
    Therefore, its limit matches its limit superior.
    Look at the theorems you learned.
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  5. #5
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    Xn is not convergent
    its bounded
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  6. #6
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    Quote Originally Posted by transgalactic View Post
    Xn is not convergent
    its bounded
    Answer:
    Quote Originally Posted by ThePerfectHacker View Post
    Let $\displaystyle X_n = x_{r_n}$.
    According to your work $\displaystyle X_n$ is convergent. TPH's $\displaystyle X_n$ is not the same as your $\displaystyle X_n$. He is talking about the subsequence which is convergent...
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  7. #7
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    $\displaystyle
    \liminf x_n \leq \limsup x_n \leq \limsup X_n = \lim X_n
    $

    why
    $\displaystyle
    limsupx_n \leq \limsup X_n
    $
    X_n is a subsequence which converges
    the definition of limsup
    is being the sup of the limit group of every subsequence

    you cant do limsup X_n
    by doing that you split this subsequence into a smaller convergent subsequences
    and take the Sup of their limits

    which has nothing to do with limsup x_n
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  8. #8
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    Quote Originally Posted by transgalactic View Post
    why
    $\displaystyle
    limsupx_n \leq \limsup X_n
    $
    If $\displaystyle A,B$ are non-empty bounded subsets with $\displaystyle A\subseteq B$ then $\displaystyle \sup (B) \leq \sup (A)$.

    Now, $\displaystyle \{ x_{r_k} | k\geq n\} \subseteq \{ x_k | k\geq n\}$.
    Therefore, $\displaystyle \sup \{ x_k | k\geq n\} \leq \sup \{ x_{r_k} | k\geq n\}$.
    Thus, $\displaystyle \lim \sup \{x_k |k\geq n\} \leq \lim \sup \{ x_{r_k}| k\geq n\}$.
    This gives, $\displaystyle \limsup x_n \leq \limsup X_n$
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