$$l_p=\{(x_i)^{\infty}_{i=1}|\sum_{i=1}^{\infty}|x_i|^p<\infty\}$$
The answer is given, but this proof makes no sense to me. If somebody could explain the logic, idea here, I would be very grateful.
$$\{e_i\}^{\infty}_{i=1}-\text{ this is the base of $l_r$ maybe???} \\\{e_i\}^{\infty}_{i=1} \subseteq l_r \subseteq l_p \\ span\{e_i\}^{\infty}_{i=1} \subseteq l_r \subseteq l_p \\ span\{e_i\}^{\infty}_{i=1}\subseteq \overline{(l_r)_{l_p}} \subseteq l_p \\ l_p\subseteq\overline{(l_r)_{l_p}} \subseteq l_p \\ l_p=\overline{(l_r)_{l_p} } \implies \text{ $l_r$ is everywhere dense in $l_p.$}$$
I am told that $\{e_i\}^{\infty}_{i=1}$ is a fundamental set in some sense.
Every $\ell^p$ space with $1\leq p\leq\infty$ contains the space $c_c(\mathbb{N})$ of sequences which are zero after finitely many terms. When $p<\infty$, $c_c(\mathbb{N})$ is a dense subset of $\ell^p(\mathbb{N})$. Since it is contained in $\ell^r(\mathbb{N})$, $\ell^r$ must be dense in $\ell^p$.
The claim is not true for $p=\infty$, because the constant sequence 1 cannot be closely approximated in $\ell^{\infty}$ by elements of $\ell^r$ for $r<\infty$.