Sequences of the form $x=\sum_{i=1}^{p}\lambda_i 1_{A_i}$ form a dense subset of $l_\infty$

Question

I have read that the set of sequences of the form

$$x=\sum_{i=1}^{p}\lambda_i 1_{A_i},$$

where $1_{A_i}$ is the characteristic function of a subset $A_i \subset \mathbb{N}$, is dense in $l_\infty$, but I cannot find a proof of that.

Can anyone tell me how to do it?

---

Motivation: What I am trying to prove is that $f \in l_\infty ^*$ assign 0 or 1 over the characteristic functions (understood as sequences of 0's and 1's) if and only of $f \in \overline{ \{e_n \} }^{\omega^*}$, where $\omega^*$ is the weak* topology over $l_\infty^*$ and $e_n$ is the basis of $l_1$ (seen in $l_1^{**}=l_\infty ^*$).

In order to prove this I come upon the other question.

Answer 1

Let $x=\{x_n\}\in\ell^\infty$ and $\varepsilon>0$ be given. We know the range of $x$ is contained in the closed ball in $\mathbb C$ of radius $\|x\|_\infty$, that is, $$ \{x_n:n\in\mathbb N\}\subset B(0,\|x\|_\infty)\equiv\{z\in\mathbb C:|z|\leq\|x\|_\infty\}. $$ Now, since $B(0,\|x\|_\infty)$ is compact, there exist $\lambda_1,\ldots,\lambda_p\in B(0,\|x\|_\infty)$ such that $$ B(0,\|x\|_\infty)\subset\bigcup_{k=1}^p B(\lambda_k,\varepsilon). $$ Now put $B_k =\{x_n: x_n\in B(\lambda_k,\varepsilon )\}$. In general, the $B_k$ will not be disjoint, so put $A_1=B_1$ and $A_k=B_k\setminus(\cup_{j=1}^{k-1}A_j)$ for $1