I am trying to understand $l^p$ spaces better and I got stuck. I showed that $l^1(\mathbb{N})$ is a subspace of $l^2(\mathbb{N})$. I also found a counterexample which shows that $l^1(\mathbb{N})$ is not a closed subspace. Could anyone show me how to find the closure of $l^1(\mathbb{N})$ in $l^2(\mathbb{N})$?
Thanks in advance!
If $(a_n)$ is in $l^2$ then $\sum |a_n|^2 < \infty$. Take $\epsilon >0$ and find an $n_0$ in $\mathbb{N}$ such that $\sum_{n\geq n_0} |a_n|^2 < \epsilon$. Then take the sequence $(b_n)$ given by $b_n = a_n$ if $n < n_0$ and $b_n=0$ otherwise. Is clear that $(b_n)$ is in $l^1$ because it is finitely supported. Observe that $||(a_n) - (b_n) ||_{l^2}^2 = \sum_{n \geq n_0} |a_n|^2 < \epsilon$.
Because $(a_n) \in l^2$ and $\epsilon >0$ were arbitrary this proves that the closure of $l^1$ in $l^2$ is $l^2$.