Let $\{x_n\}$ be a bounded sequence and $s=\sup\{x_n|n\in\mathbb N\}.$ Show that if $s\notin \{x_n|n\in\mathbb N\}, $then there exists a subsequence .

Question

Let $\{x_n\}$ be a bounded sequence and $s=\sup\{x_n|n\in\mathbb N\}.$ Show that if $s\notin \{x_n|n\in\mathbb N\}, $then there exists a subsequence convereges to $s$.

$s-1$ cannot be the upperbound and $s\notin \{x_n|n\in\mathbb N\}, \exists n_1\in \mathbb N:n_1\ge1:s-1<x_{n_1}<s.$

$s-\frac{1}{2}$ cannot be the upperbound and $s\notin \{x_n|n\in\mathbb N\}, \exists n_2\in \mathbb N:n_2\ge1:s-\frac{1}{2}<x_{n_2}<s.$

$\dots$

$s-\frac{1}{k}$ cannot be the upperbound and $s\notin \{x_n|n\in\mathbb N\}, \exists n_k\in \mathbb N:n_k\ge1:s-\frac{1}{2}<x_{n_k}<s.$

$\dots$

My doubt

I can create a sequence of a number like this. How do we guarantee that we can choose $n_2>n_1$ such that $s-\frac{1}{2}<x_{n_2}<s$ hold? similarly for other cases. If we can show this then required subsequence would be $\{x_{n_k}\}$.

Answer 1

One can do it like looking for $n_{k+1}\geq 1$ such that $\max\left\{s-\dfrac{1}{k+1},x_{1},...,x_{n_{k}}\right\}<x_{n_{k+1}}<s$.

Answer 2

Using the definition of supremum and the assumption that $s\notin \{x_n\}$, show that for any $\epsilon > 0$, there are infinitely many $n$ such that $s-\epsilon < x_n < s$. Having done so, find an increasing sequence $n_1 < n_2 < n_3 < \dots,$ such that $s-1/n_i < x_{n_i} <s$.