Consider $c_{00}$ as a subspace of $(\ell^p,\|\cdot\|_p)$. Show that the closure of $(c_{00},\|\cdot\|_1)$ is $\ell^1$, closure of $(c_{00},\|\cdot\|_2)$ is $\ell^2$ and closure of $(c_{00},\|\cdot\|_\infty)$ is $c_0$.
$$c_{00}=\{x=(x(1),x(2),\ldots): x(j)=0\ \forall j \ge j_x, \exists j_x \in \mathbb{N}\}$$
So I create a set of sequences like this: $$ \begin{align} x_1 & =(x(1),0,0,0,\ldots) \\ x_2 & =(x(1),x(2),0,0,\ldots) \\ & \,\,\,\vdots \\ x_n & =(x(1),x(2),\ldots,x(n),0,0,\ldots) \end{align} $$ where the $$x=(x(1),x(2),\ldots,x(n),\ldots)$$ is the parent sequence which I pick in the various spaces. I think that these sequences will do the work for me.
For (a) $(c_{00},\|\cdot\|_\infty)$
let $$x=(x(1),x(2),\ldots,x(n),\ldots) \in c_0$$ So for any $\varepsilon > 0$ there exists a $n_0 \in \mathbb{N}$ such that $|x(n)| < \frac{\varepsilon}{2}$ for all $n \ge n_0$. Now for all $n \ge n_0$ , $$\|x_n-x\|_\infty=\sup\{|x(m)|, m > n_0\} \le \frac{\varepsilon}{2} < \varepsilon$$
for (b) , $(c_{00},\|\cdot\|_p)$
let $$x=(x(1),x(2),\ldots,x(n),\ldots) \in \ell^p$$ Now $$\|x_n-x\|_p=\left(\sum_{i=n+1}^\infty|x_i|^p \right)^{\frac{1}{p}}$$
Now since $x \in \ell^p$, the series $$\left(\sum_{i=1}^\infty |x_i|^p \right)^{\frac{1}{p}} < \infty$$ and hence the tail of the series goes to $0$. Thus there is a $n_0 \in \mathbb{N}$ such that for all $n \ge n_0$ $$\left(\sum_{i=n}^\infty |x_i|^p \right)^{\frac{1}{p}} < \epsilon $$
And we are done. Thanks for the help!!
