Interior and closure of $l^1 (\mathbb{N})$ in $ l^{\infty}$? Let $$ l^{\infty} = \left\{ (x_n)_{n \in \mathbb{N}} \in \mathbb{R}^{\mathbb{N}} \mid (x_n)_{n \in \mathbb{N}} \ \text{is bounded} \right\} $$ Here $\mathbb{R}^{\mathbb{N}}$ denotes the space of all sequences in $\mathbb{R}$. On $l^{\infty}$ we consider the metric $$ d_{\infty} ((x_n)_n, (y_n)_n) = \sup \left\{ | x_n - y_n | \mid n \in \mathbb{N} \right\}. $$  Now consider the subset $l^1({\mathbb{N}})$ of the sequences whose associated series are absolute convergent: $$ l^{1} (\mathbb{N}) = \left\{ (x_n)_{n \in \mathbb{N}} \in \mathbb{R}^{\mathbb{N}} \mid \sum_{n = 0}^{\infty} | x_n | \ \text{converges} \right\}. $$ 
Question: What is the interior and closure of $l^1 (\mathbb{N}) $ in the metric space $(l^{\infty} (\mathbb{N}), d_{\infty} ) $?
My attempt: I think the interior of $l^1 (\mathbb{N})$ is just the set itself, and the closure is empty. Since $l^1(\mathbb{N})$ consists of sequences whose associated series are absolute convergent, we know that the terms far enough in the sequence will tend to zero. Lets says I take a sequence $(x_n)_n$ in the interior of $l^1 (\mathbb{N})$. Then there exists a $\delta > 0 $ such that the open ball $B((x_n), \delta)$ lies within $l^1 (\mathbb{N})$. Should I now show that if I take another sequence $(y_n)_n \in l^1 (\mathbb{N})$, this sequence will also lie in the open ball?
I'm not sure how to prove my assertions in a rigorous manner.
 A: (1). $l^1$ is not closed in $l^{\infty}:$ Let $x_n=(x_{n,i})_{i\in N}$ where $x_{n,i}=1/i$ for $i\leq n,$ and $x_{n,i}=0$ for $i>n.$ Let $y=(1/n)_{n\in N}.$ Then each $x_n\in l^1$ and $y\in l^{\infty}$  \ $l^1$ but in the $\sup $ norm we have $\|y-x_n\|=1/(n+1).$ So $y\in \overline {l^1}$  \ $l^1.$
(2). The closure of $l^1$ in $l^{\infty}$ is not equal to $l^{\infty}:$ Let $z=(z_n)_{n\in N}$ where  $z_n=1$ for every $n\in N.$  Now  $x=(x_n)_{n\in N}\in l^1\implies\lim_{n\to \infty}x_n=0\implies \|z-x\|\geq  \lim_{n\to \infty}|z_n-x_n|= 1.$ So $z\not \in \overline {l^1}.$
(3). If $X$ is a real or complex normed linear space and $Y$ is a vector subspace of $X$ with $Y\ne X$ then int$_X(Y)=\phi.$ Because otherwise, take a non-empty open ball $B$ centered at $y,$ with $B\subset Y.$ Then $(-y)\in Y,$ and $C=(-y)+B=\{-y+z:z\in B\}$ is a non-empty open ball centered at $0,$ with $C\subset  Y.$ But every $x\in X$ is a scalar multiple of a member of  $C,$ and hence a member of $Y.$  (4).By (2)and (3) we have int$(l^1)\subset$ int $(\;\overline {l^1}\;)=\phi.$ 
