I am trying to show $\ell^p$ is dense in $\ell^q$, where p< q so i need to show that any $\epsilon > 0$ for any $(x_i)\in\ell^q$ there is $(y_i) \in \ell^p$ such that $d((x_i),(y_i)) < \epsilon$. Let be $(x_i) \in l^q$ and $\epsilon$ be orbitrary. We know $\sum_{i=1}^\infty |x_i|^q <\infty$ this means $\exists$ n such that $\sum_{i=n+1}^\infty |x_i|^q < {(\epsilon)^p \over 2^p}$ .
Consider $y=(y_1,y_2, . . .,y_n,0,0,. . .)$ and let $(y_i) \in \ell^p$ and each $y_i$ is rational. Since the rationals are dense in $\Bbb{R}$ , for each $x_i$ there is rational $y_i$ close to it. Hence we can find a $y \in \ell^p$ satisfying
$\sum_{i=1}^\infty |x_i-y_i|^q< {\epsilon^p\over 2^p}$. It follows that $d(x,y)=(\sum_{i=1}^\infty |x_i-y_i|^q)^{1\over p} < (\sum_{i=1}^n |x_i-y_i|^q + \sum_{i=n+1}^\infty |x_i|^q)^{1\over p} < \epsilon$
Is it true? If it is not true, can you show me the way to do it please? Thank you for your help.
As Nate Eldredge and copper.hat said: the proof works, but there's no need to go to the rationals. You can just take $y_i = x_i$ for $1 \le i \le n$.
The idea of proof is to show that sequences with finitely many nonzero elements are dense in $\ell^q$. This is sufficient for conclusion, because such sequences are contained in every $\ell^p$.