I go through a proof of the following.
Let $(\ell_1,d)$ be the metric space of all sequences $x = (\xi_i)_{i \in \mathbb{N}}$ with $\sum_{i=1}^{\infty} |\xi_i| < \infty$ and the metric $ d(x,y) = \sum_{i=1}^{\infty} |\xi_i - \eta_i|, \qquad x = (\xi_i), y = (\eta_i). $
Theorem: A subset $M$ of $l_1$ is totally bounded (pre-compact) iff
(i) There is a $K > 0$ with $\sum_{i=1}^{\infty} |\xi_i| \le K$ for all $x = (\xi_i) \in M$;
(ii) $\forall \varepsilon > 0 \exists n_0 \in \mathbb{N} \forall x = (\xi_i) \in M: \sum_{i=n_0}^{\infty} |\xi_i| < \varepsilon$.
The proof goes like this, let $M$ be a subset such that (i) and (ii) hold and let $x = (x_i)$ be a sequences in $\ell_1$ (i.e. a sequence of sequences), we show that it has a sub-sequence $x' = (x'_i)$ with $x'_i = (\xi^{(i)}_j)$ which is a Cauchy-Sequence. So let $\varepsilon > 0$ be given, then select $n_0$ such as in (ii), then for $n,m > n_0$ \begin{align*} d(x_n, x_m) &= \sum_{i=1}^{\infty} | \xi^{(n)}_i - \xi^{(m)}_j | \le \sum_{i=1}^{\infty} | \xi^{(n)}_i | - | \xi^{(m)}_j | \le 2\varepsilon \end{align*} (end of proof)
I dont understand the last steps, why is this sum smaller than $2\varepsilon$?.