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Given a bounded sequence $\{a_n\}_{n=0}^{\infty}$ and the sequence $\{b_m\}_{m=0}^{\infty}$ defined by: $$b_m=\sup \{a_m, a_{m+1},...\}=\sup \{a_n|n \ge m\}\;.$$

Prove that there exists $L=\lim_{m\to\infty} b_m$ and that $L$ is a subsequential limit of $\{a_n\}_{n=0}^{\infty}$. ( is it the limsup of $a_n$?)

Thank you very much.

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1 Answer 1

Note that $b_n$ is a monotonically non-increasing and bounded sequence because $a_n$ is bounded. Then it is classical that the limit $L$ of $b_n$ exists. Now, given $ \epsilon \ > 0$ there exists $ b_n \in (L - \epsilon/2, L +\epsilon/2) $ and by definition there exists $ a_{n_m} \in (b_n - \epsilon/2, b_n + \epsilon/2)$. Hence the subsequence $ a_{n_m}$ converges to $L$.

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In fact, the sequence $(b_n)_n$ is nonincreasing, that is, $b_{n+1}\leqslant b_n$ for every $n$. (To see this, note that $b_n=\sup\{a_n,b_{n+1}\}$ for every $n$.) –  Did Apr 30 '12 at 18:00
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