Separability of $l^{p}$ spaces

Question

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!

Answer 1

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.