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?

  • $\begingroup$ 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

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 $(x_{n_i}) \to S$.

  • 1
    $\begingroup$ I agree with you but is the method I gave ok? $\endgroup$ – z23 Mar 10 '14 at 18:39
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.