Exterior of a set in metric space Consider the metric space $(\ell^\infty(\mathbb{N}),d_\infty)$ where  $\ell^\infty (\mathbb{R}=\{ (x_n)_n \in \mathbb{R}^{\mathbb{N}} \text{ where } (x_n)_n \text{ is bounded}\}$ and $d_\infty((x_n)_n,(y_n)_n)=\sup \{|x_n-y_n | $ where $n \in \mathbb{N} \}$. Now consider the subset $\ell^1(\mathbb{N})=\{ (x_n)_n \in \mathbb{R}^{\mathbb{N}}$ where $\sum_{n=0}^\infty|x_n|$ converges$\}$. What is the interior, exterior and closure of $\ell^1(\mathbb{N})$?
My reasoning for the interior goes as follows: $(\ell^1(\mathbb{N}))^\circ= \emptyset$. Indeed, take a random $x \in \ell^1(\mathbb{N})$. Then there exists for every $\varepsilon >0$ an $y \in \ell^\infty(\mathbb{N}) \setminus \ell^1(\mathbb{N})$ so that $d_\infty(x,y)< \varepsilon$. Take for example $y=x+ \frac{\varepsilon}{2}$. (The terms of $x$ go to zero since its series converges absolutely and so the terms of $y$ go to $\frac{\varepsilon}{2} >0$ so that certainly $y \in \ell^\infty(\mathbb{N}) \setminus \ell^1(\mathbb{N})$.)
I am not sure how to start on the exterior/ closure. Can someone help? Thanks!
 A: Let us define: 
$$c_0 := \{ (x_n)_{n \in \Bbb{N}} \in l^\infty(\Bbb{N}) | \lim\limits_{n \to \infty} x_n = 0 \}$$
$$c_{00} := \{ (x_n)_{n \in \Bbb{N}} \in l^\infty(\Bbb{N}) |\exists N \in \Bbb{N}\colon \forall n \ge N \colon x_n = 0\} $$
It is an easy exercise to show $c_{00} \subset l^1(\Bbb{N}) \subset c_0$ and to prove that $c_{00}$ is dense in $c_0$ with respect to the given metric. By taking the closure of all sets in above chain you get: 
$$ c_0= \overline{c_{00}} \subset \overline{l^1(\Bbb{N})} \subset \overline{c_0} = c_0 $$
Therefore one has $\overline{l^1(\Bbb{N})} = c_0$. That means the closure of $l^1$ is given by all sequences converging to $0$.
Now Let $(M,d)$ be a metric space and $S \subset M $ then it holds that the exterior of $S$ is given by $\text{ext}(S) = M\setminus \overline{S}$.
Applying this identity we get: 
$$ \text{ext}(l^1(\Bbb{N})) = l^\infty(\Bbb{N}) \setminus \overline{l^1(\Bbb{N})} = l^\infty(\Bbb{N}) \setminus c_0 $$
Our metric space is here $M = l^\infty(\Bbb{N}) $ and $S$ is choosen as $S = l^1(\Bbb{N})$.
For reference on the used identity see for example  https://en.wikipedia.org/wiki/Exterior_(topology) under Equivalent Definitons. For information on the above facts about $c_0,c_{00}$ you might look into an arbitary book about functional analysis.
A: [1]. The closure of $l^1$ in $l^{\infty}$ is the set of sequences that converge to $0$:
(i). If $x=(x_n)_n$ converges to zero then for $\epsilon >0$ take $n_0\in \mathbb N$ such that $n>n_0 \implies |x_n|<\epsilon.$ Let $y_i=x_i$ for $i\leq n_0$ and $y_i=0$ for $i>n_0.$ Then $y=(y_i)_i\in l^1$ and $\|x-y\|_{\infty}\leq \epsilon.$
(ii).  If $x=(x_n)_n\in l^{\infty}$ does not converge to $0$ then take $r>0$ such that $\{n:|x_n|>r\}$ is an infinite set. Then any $x'=(x'_n)_n$ with $\|x'-x\|_{\infty}<r/2$ is not in $l^1$ because there are infinitely many $n$ for which $|x'_n|>r/2. $
[2]. Regarding the interior of $l^1$ in $l^{\infty}:$ More generally if $V$ is a normed vector space and $W$ is  vector sub-space of $V$ whose interior, in $V,$ is not empty, then $W=V.$
For if $r>0$ and $x\in V$ and the open ball $B(x,r)=\{y\in V:\|y-x\|<r\}$ is a subset of $W$ then (obviously) $x\in W,$ so $W\supset \{-x+y:y\in B(x,r)\}=B(0,r).$ Now for  $0\ne z\in V$ let $z'=\frac {z}{\|z\|}\cdot \frac {r}{2}.$ Then $z'\in B(0,r)\subset W$ so $z=z'\cdot \frac {2\|z\|}{r}\in W.$ (And, pedantically, also $0\in W.$) 
