Let the set be given like in the following manner: $$\{x_n: n\in\mathbb N\}\subset \mathbb{R^n}$$ $$l^2=\left\{\{x_{n}\}_{n=1}^{\infty}\,\Big|\, \sum_{n=1}^{\infty}|x_n|^2<\infty\right\}.$$
Prove that there exist closed and bounded sets, such that are not compact. (notebook answer) \begin{align} E&=\{e_n\mid n \in \mathbb{N}\} \\ e_1&=(1,0,0,\ldots,0,\ldots) \\ e_2&=(0,1,0,...,0,...) \\ &\;\;\vdots \\ e_n&=(0,0,0,\ldots,1,\ldots) \end{align} with $\|e_n \|=1, n \in \mathbb{N}$.
This set is bounded , because it is always on a unit sphere , but this sequence isn't compact because the members are always on a fixed distance:$$\|e_n-e_m\|=\begin{cases}0,& n=m \\{} \\ \sqrt{2},& n\neq m \end{cases}$$
What is the reasoning behind this predication?
Hint: Consider an open cover of $E$ consisting of balls of radius, say, 1.3, centered at the points $e_n$, i.e. $\{B(e_n, 1.3) : n = 1, 2,\dots\}$. Does it have a finite subcover? (For that matter, does it have any proper subcover?)