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

My proof:

For existence, we know that from the Bolzano-Weirstrass Theorem, that such a sequence does exist. Also we have that $m \le x_k \le M$ for all $k \in N$. So $m$ is a lower bound for $\{x_k:k \ge n\}$. So that means it must be a lower bound for the subsequence. Hence $m \le \sup\{x_k:k \ge n\}$, which implies that $m - \epsilon \le \inf{s_n} - \epsilon$, which is equivalent to $m-\epsilon \le S-\epsilon$. Hence we have for $n_k \ge k$, $S-\epsilon \le x_{n_k}$, which is equivalent to $-\epsilon \le x_{n_k} - S$, and since $\epsilon$ is arbitrary, this implies convergence to $S$.

Is my proof correct?

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  • $\begingroup$ What is your question? $\endgroup$ – 6005 Mar 10 '14 at 17:40
  • $\begingroup$ whoops my question is, is the proof correct? $\endgroup$ – z23 Mar 10 '14 at 17:49
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Let $y_k= \sup\{x_n:n\ge k\}$ and $S$ be the limit superior. You need to show that given $\alpha$ and $\beta$ such that $\alpha<S<\beta$, the set $\{k\in \mathbb{N}: \beta< x_k\}$ is finite and $\{k\in\mathbb{N}:\alpha<x_k\}$ is infinite.

After you do this. You can define $n_i=\min\{n\in\mathbb{N}: | x_n-S|\le 2^{-i} \text{ and } n\not=n_k \text{ for all } k<i \}$ and so the subsequence $(x_{n_i}) \to S$.

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    $\begingroup$ I agree with you but is the method I gave ok? $\endgroup$ – z23 Mar 10 '14 at 18:39
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    $\begingroup$ @z23: I think there is a flaw in the last point "Hence we have for $n_k \ge k$, $S-\epsilon \le x_{n_k}$"of course, there are infinitely many such $x_{n_k}$ but this does not means that it holds for all. $\endgroup$ – Jose Antonio Mar 10 '14 at 18:45

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