Proving relative compactness of a set on $\ell_{p}$ space Assume $M \subset \ell_p$, $p \geq 1$ is bounded and for every $\varepsilon > 0$ there exists such $N_{\varepsilon} \in \mathbb{N}$, that for every $x = (x_i) \in M$ the following holds:
$$ \sum_{i = N_{\varepsilon}}^{\infty}  | x_i | ^{p} < \varepsilon^{p}$$
I'm trying to show that the set $M$ is relatively compact in $\ell_p$.
Since $\ell_p$ is complete it should suffice showing that for every $\varepsilon >0$ there exists a finite $\varepsilon$-net ($M_{\varepsilon}$) of the set $M$. 
How could we construct such set? 
Current thoughts: It seems that the main idea here is to construct the first  $N_{\varepsilon/2}$ elements of $x_\varepsilon \in M_{\varepsilon}$, such, that for any $x \in M$
$$ d(x_\varepsilon, x)^p = \sum_{i = 1}^{N_{\varepsilon/2}-1}  |x_{\varepsilon,i}- x_i | ^{p} + \sum_{i = N_{\varepsilon/2}}^{\infty}  |x_i | ^{p} < \varepsilon^{p}/2 + \varepsilon^{p}/2$$
Following the proof of separability of $\ell_p$, it's clear that in a similar manner a countable set of finite sequences with rationals can be constructed. However, here we require $M_\varepsilon$ to be a finite set.
What would be the best way proceed in this case? Or maybe an alternative approach would be better? 
 A: Let $\varepsilon > 0$. Define $N = N_{\varepsilon 2^{-\frac1p}}$ such that $$\sum_{i=N}^\infty |x_i|^p < \frac{\varepsilon^p}2, \forall x = (x_i) \in M$$
$M$ is bounded so $\exists R > 0$ sucht that $\|x\|_p \le R, \forall x \in M$. The closed ball $\overline{B}(0, R) \subseteq \mathbb{C}$ is totally bounded so for $r = \frac{\varepsilon}{(2(N-1))^{1/p}}$ there exists a finite set $A \subseteq \overline{B}(0, R)$ such that
$$\overline{B}(0, R) \subseteq \bigcup_{a \in A} B(a, r)$$
Define $$S = \{(a_1, a_2, \ldots, a_{N-1}, 0, 0 ,\ldots) : a_i \in A\} \subseteq \ell^p$$
$S$ is a finite set, namely $|S| = |A|^{N-1}$. We claim that $M \subseteq \bigcup_{y \in S} B(y, \varepsilon)$. 
Indeed, let $x = (x_i) \in M$. We have $|x_i| \le \|x\|_p \le R$ so $x_i \in \overline{B}(0, R), \forall i \in \mathbb{N}$. In particular, for $i \in \{1, \ldots, N-1\}$ there exists $a_i \in A$ such that $x_i \in B(a_i, r)$.
Let $y = (a_1, \ldots, a_{N-1}, 0, 0, \ldots )\in S$. We have
$$\|x-y\|_p^p = \sum_{i=1}^{N-1}|x_i - a_i|^p + \sum_{i=N}^\infty |x_i|^p < (N-1)r^p + \frac{\varepsilon^p}2 < \frac{\varepsilon^p}2 + \frac{\varepsilon^p}2 = \varepsilon^p$$
so $x \in B(y, \varepsilon)$.
We conclude that $M$ is relatively compact.
