Showing a set is dense in $l^1$ I have been studying functional analysis and I came across this question: Show that $$F := \{(s_n)\in l^1 : (ns_n)\notin l^{\infty}\}$$ is dense in $l_1$.
So here is my approach and I would like some feedback if possible: $(e_n) \in l^1$ (a sequence with unit value in the nth position) is certainly part of this set as $(ne_n)$ is unbounded. We know already that the span of $(e_n)$ is dense in $l^1$. Every other sequence in F can be represented by the span of $(e_n)$. Hence, $(e_n)$ is dense in F itself. Hence, F is dense in $l^1$.
Is this correct? I am sorry that I can be very vague in structuring my arguments (I am new to mathematics) and was wondering if anyone could help me with structuring this better or correct me if I am wrong somewhere.
 A: Given $\epsilon>0$ and a sequence $(s_n)\in\ell^1$.
Let $u_n:=\begin{cases}\frac{m}{2^m}&n=2^m\\0&o/w\end{cases}$; and let $\epsilon_n:=\frac{\epsilon u_n}{4}$.
Then $(\epsilon_n)\in\ell^1$ with $\|(\epsilon_n)\|_1=\frac{\epsilon}{2}$, since $\sum_nu_n=\sum_m\frac{m}{2^m}=2$.
Consider $b_n:=s_n\pm\epsilon_n$, with the sign chosen to be the same as that of $s_n$;
then $(b_n)\in\ell^1$ since $\|(b_n)\|_1\le\|(s_n)\|_1+\|(\epsilon_n)\|_1<\infty$;
and $(nb_n)\notin\ell^\infty$ since $|nb_n|=|ns_n|+n\epsilon_n\ge n\epsilon_n$, and $nu_n=\begin{cases}m&n=2^m\\0&o/w\end{cases}$ is unbounded.
Hence $$\|(s_n)-(b_n)\|_1=\|(\epsilon_n)\|_1=\frac{\epsilon}{2}$$
This means that every ball around $(s_n)\in\ell^1$ has a sequence $(b_n)\in F$ within it.
A: I think it is easiest to prove this directly. Fix some $\varepsilon>0$ and some $(x_n)\in\ell^1$. We know there must exist some $K\in\mathbb N$ such that $\sum_{n=K}^\infty|x_n|<\varepsilon/2$. Now note that there exists an infinite set $A\subset \mathbb N$ so that
$$\sum_{n\in A} 1/\sqrt n<\varepsilon/2.$$
Now define some $(s_n)$ by $s_n=x_n$ for all $n\leq K$. For all $n\geq K$ let
$$s_n=\begin{cases}1/\sqrt n &  \text{if}~ n\in A\\ 0 & \text{if}~n\notin A.\end{cases}$$
Clearly $(s_n)\in\ell^1\cap F$, and $\|(s_n)-(x_n)\|_1<\varepsilon$. As $(x_n)$ and $\varepsilon$ are arbitrary it follows that indeed $F$ is dense in $\ell^1$.
