Separability of $l^{p}$ spaces How can I prove that the space $l^{p}$ equipped with the norm (for $x=(x_{n}) \in l^{p})$:
$\|x\|_{p}=(\displaystyle\sum_{n}|x_{n}|^{p})^{1/p}$
Is a separable space? (i.e. showing that there is a countable dense set in $l_{p}$). I saw a proof in which they started with first letting $x=(x_{n})$ be an element of $l^{p}$ and for any $\epsilon >0$, then they chose $N$ such that:
$(\displaystyle\sum_{n>N}|x_{n}|^{p})<(\frac{\epsilon}{2})^{p}$
However, I'm not sure where does this come from. As well, I'm not sure how to proceed from this step.
Thank you for your help!

The idea here is to see that the set of finite sequences is dense in $l^p$ (for $p < \infty$)
So we will approach $x$ by the sequence $x^{(n)}$, with $x^{(n)}_i = x_i$ if $i\leq n$ and  $x^{(n)}_i = 0$  if $i > n$
This gives you 
$$\| x-  x^{(n)} \|_p^p = \sum_{k>n} |x_k|^p$$
As the serie $\sum |x_k|^p$ converge, the tail converge to zero, so you have
$$\| x-  x^{(n)} \|_p^p \to 0$$
The second step is to approximate each $x^{(n)}$ by elements of a countable subset. If it's $l^p(\mathbb{R})$, the set of the finite sequences at value in $\mathbb{Q}$ works well:


*

*it's countable, because it's the countable union of countable sets

*the $x^{(n)}_i$ can be uniformly approximated by a sequence $(y^{(n,i)}_{k})_{k\in\mathbb{N}}$ of rationals, so each $x^{(n)}$ can be approximated by a sequence of sequences $y^{(n)}$ such that $y^{(n)}_i=y^{(n,i)} $


So 
$$\|x- y^{(n,i)} \| \leq \underbrace{\|x- x^{(n)} \|}_{\to 0}+ \underbrace{\| x^{(n)}-y^{(n,i)} \|}_{\to 0} $$
and you have the density.
