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Thread: Proving a subsequence converges to S

  1. #1
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    Proving a subsequence converges to S

    Let $\displaystyle (x_n)$ be a bounded sequence for each $\displaystyle n \in N$ let $\displaystyle s_n:=$sup{$\displaystyle x_k:k$>$\displaystyle n$} and $\displaystyle S=inf${$\displaystyle s_n$}. Show that there exists a sub sequence of $\displaystyle (x_n)$ that converges to S.

    Ok so clearly $\displaystyle s_1$>$\displaystyle s_2$>....>$\displaystyle s_n$.

    Since it is decreasing and bounded then $\displaystyle s_n$ converges to $\displaystyle S$. Futhermore there exists a subsequence of $\displaystyle s_n$ that is convergent that much also converge to S. My only problem is I don't know how to relate this back to $\displaystyle (x_n)$
    Last edited by hockey777; Sep 24th 2008 at 01:02 PM.
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  2. #2
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    I take that back, there is still need for help.
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  3. #3
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    Quote Originally Posted by hockey777 View Post
    Ok so clearly $\displaystyle s_1$>$\displaystyle s_2$>....>$\displaystyle s_n$.
    This does not make sense, $\displaystyle s_n$ are sets, how can they be "larger" or "smaller"?
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  4. #4
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    Quote Originally Posted by ThePerfectHacker View Post
    This does not make sense, $\displaystyle s_n$ are sets, how can they be "larger" or "smaller"?

    no $\displaystyle s_1$ is the first position in the sequence $\displaystyle s_n$, but I see where you think that as I messed up on my original post, I fixed it.
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