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

Question

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

Proof:

The normed linear space $l_p$ ,$1\leq 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 $(x_n)_{n\geq 1}=x\in l^p,1\leq p<\infty,$ 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 $n\in\Bbb{N}$ such that,
$\sum_{i=n}^{\infty}|x_i|^p<\epsilon^p/2.$
Since rationals are dense in $\Bbb{R}$, therefore for given $\epsilon>0$,
there exist rational coordinates $y_1,y_2,…,y_n$ such that,

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

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.

Thanks @Kavi Rama Murthy!

Answer 1

The idea is correct but some changes are necessary in the presentaion:

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

$$|x_i-y_i|<(\frac{1}{2n})^{p}\epsilon.$$

You should say there exist rational numbers $y_1,y_2,...,y_n$ such that

$$|x_i-y_i|<(\frac{1}{2n})^{1/p}\epsilon.$$ and hence

$$\sum_{i=1}^{n-1}|x_i-y_i|^p<\frac{\epsilon^{p}}{2}.$$

There is confusion beteeen $n$ and $N$ . You could drop $N$ and use $n$ everywhere.