functional analysis – The normed linear space $l_p$ ,$1leq p

The normed linear space $l_p$ ,$1leq p<infty $ is separable.

Proof:

The normed linear space $l_p$ ,$1leq p<infty $ is separable. The set $S$ of all sequence having finitely many non zero rationals hence $S$ is countable.

Now we claim that $S$ is dense subset of $X$.

Let $(x_n)_{ngeq 1}=xin l^p,1leq p<infty,$ then $sum_{i=1}^{infty}|x_i|^p<infty$. Thus this series is convergent hence tail of this series is also convergent i.e., for given $epsilon>0$, there exist $ninBbb{N}$ such that,
$sum_{i=N}^{infty}|x_i|^p<epsilon^p/2.$

let $y_1,y_2,…,y_n$ be non zero rational coordinates, Since rationals are dense in $Bbb{R}$, therefore for given $epsilon>0$,

$$|x_i-y_i|<(frac{1}{2n})^pepsilon.$$
Therefore,
$$sum_{i=1}^{N-1}|x_i-y_i|^p<frac{epsilon}{2}.$$

Now $d(x,y)=||x-y||=(sum_{i=1}^infty|x_i-y_i|^p)^frac{1}{p}=(sum_{i=1}^{N-1}|x_i-y_i|^p+sum_{i=N}^{infty}|x_i|^p)^frac{1}{p}<(frac{epsilon^p}{2}+frac{epsilon^p}{2})^frac{1}{p}=epsilon.$ Hence any open ball containing $x$ contains some points of $S$, thus $S$ is dense in $l_p$ space . This completes the proof.

Is this proof correct?