For each $n\in\mathbb{N}$ let $e_{n}=(x_{k})_{k\in\mathbb{N}}$ with $x_{n}=1$ and $x_{i}=0$ for all $i\neq n$. Then $A=\{e_{n}:n\in\mathbb{N}\}\subset\mathcal{l}^{2}$ and $d(e_{n},e_{m})=\lVert e_{n}-e_{m}\rVert=\sqrt{2}$. Show that, if $A\subset X\subset\mathcal{l}^{2}$, then $X$ is not pre-compact( or totally bounded) on $(\mathcal{l}^{2},\lVert \rVert)$.
By definition of totally bounded, for all $\epsilon>0$, we can a descomposition $M=X_{1}\cup\cdots\cup X_{n}$ of $M$ as the reunion of finite number of subsets, each of diameter $<\epsilon$, so how prove this, any idea.. thanks!
If $X$ is precompact, then its closure $\overline{X}$ is compact. As $\ell^2$ is a metric space, this implies that $\overline{X}$ is sequentially compact. But the sequence $\{e_n\}$ in $\overline{X}$ has no convergent subsequence because $||e_n-e_m||_2=\sqrt{2}$ for all $n\neq m$, which is a contradiction.
For the second part, I claim that no ball of radius $\frac{1}{2}$ can contain both $e_n$ and $e_m$ for $n\neq m$. Indeed, if $B$ has center $x$ and radius $\frac{1}{2}$, then for any points $y,z\in B$, $$ ||y-z||_2\leq ||x-y||_2+||x-z||_2<\frac{1}{2}+\frac{1}{2}=1<\sqrt{2} $$ Therefore $X$ cannot be covered by finitely many balls of radius $\frac{1}{2}$, so is not totally bounded.