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

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

Proof:

The normed linear space $$l_p$$ ,$$1leq p is separable. The set $$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 $$(x_n)_{ngeq 1}=xin l^p,1leq p then $$sum_{i=1}^{infty}|x_i|^p. 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

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

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?