interior and closure in metric spaces let say we have $(\ell^{1}(\Bbb{N}),d_{1})$ as a metric space with $d_{1}((x_{n})_{n},(y_{n})_{n})=\sum_{n=0}^{\infty}|x_{n}-y_{n}|$. If $$D=\left\{x \in \ell^{1}(\Bbb{N}) \,\,\Big|\, \sum_{n=1}^\infty n|x_{n}|<\infty \right\}$$
I'm looking for the interior of $D$ and the closure of $D$.
I thought that the interior of $D$ was empty.
This is my attempt to prove it:
Let's say that the interior isn't empty
take an random $x$ element of $l^{1}(N)$ that belongs to the interior of $D$. Then there is a $\delta >0$ so that $B(x,\delta)$ belongs to $D$.
Now take $y=\left(\frac{x_n}n\right)_n$ then $y$ isn't an element of $D$ but $d_{1}(x,y)<\delta$.
So the interior must be empty.
Now i'm not sure if my proof is correct and i find this kind of excercises really difficult. I hope someone can help me to explain this to me and help me to do it right.
For the exterior i thougt that it was just D but i can't even start to prove it.

You are correct that $\operatorname{Int} D = \emptyset$ but for your $y$ I can't see why would $d_1(x,y) < \delta$.
The idea is similar, though. Assume that $B(x,\delta) \subseteq D$. Notice that $D$ is a vector subspace of $\ell^1$, meaning it is closed under addition and scalar multiplication. Pick some $y \in \ell^1 \setminus D$ such as $y = \left(\frac1{n^2}\right)_n$ and notice that
$$d_1\left(\frac{y}{\|y\|_1}\frac{\delta}2 - x,x\right) = \left\|\frac{y}{\|y\|_1}\frac{\delta}2\right\| = \frac\delta2 < \delta$$
so $z := \frac{y}{\|y\|_1}\frac{\delta}2 - x$ is contained in $B(x,\delta) \subseteq D$. Now we have
$$y = \frac{2\|y\|_1}{\delta}(x+z) \in D$$
because $D$ is a subspace. But this is a contradiction since $y \notin D$. Therefore interior of $D$ is empty.
Regarding the closure, notice that $D$ contains the set $c_{00}$ of all finitely supported sequences in $\ell^1$ and recall that $c_{00}$ is dense in $\ell^1$. Threfore
$$\overline{D} \supseteq \overline{c_{00}} = \ell^1$$
so the closure is $\overline{D} = \ell^1$.
