0
$\begingroup$

Let $(x_n)$ be a bounded sequence, and for each $n\in\mathbb{N}$ let $s_n=\sup\{x_k:k\geq n\}$ and $S=\inf\{s_n\}$. I need to show that there exists a subsequence of $(x_n)$ that converges to $S$.
I cannot proceed in the question as I am not sure what implication $s_n$ and $S$ have in terms of series convergence

$\endgroup$
0
$\begingroup$

Note that $s_n$ is monotonically decreasing to $S$ as $n \rightarrow \infty$. Since $s_{n}$ is the supremum of $\{x_k : k \geq n\}$, we can inductively pick $x_{n_i}$ with $n_{i+1} > n_{i}$ such that $x_{n_i} > S + 1/i$. It follows that $x_{n_i}$ goes to $S$ as $i \rightarrow \infty$.

$\endgroup$

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.