Fundamental set in the space of bounded sequences

Question

Definition: Set $S$ fundamental set in Banach space $X$ if $\overline{Lin(S)}=X$.

If $e_n=(0,\ldots ,0,1,0,\ldots)$ is a sequence that has $0$ everywhere, except on the $n$-th place and $e=(1,1,1,\ldots)$ is a constant sequence, then the set $S=\{e_n|n\in \mathbb{N}\}\cup\{e\}$ is fundamental in the space $c$ of all convergent sequences, but it's not fundamental in the space $l^{\infty}$ of all bounded sequences. Why? Is there a fundamental set in $l^{\infty}$, other than the $l^{\infty}$ itself? If yes, what is it?

Answer 1

Let $(a_n) \in c$ and $a=\lim a_n$. Then $\|(a_n)- \sum\limits_{k=1}^{N} (a_k-a)e_k -ae\| \leq \epsilon$ for any $N$ such that $|a_n-a| <\epsilon$ for all $n \geq N$. Hence $S$ is fundamental in $c$.

Consider the sequence $(a_n)=(1,-1,1,-1,...)$. If there exist $a,N$ and $c_i$'s such that $\|(a_n)- \sum\limits_{k=1}^{N} c_ke_k -ae\| \leq \epsilon$ then we get $|(-1)^{n}-a|\leq \epsilon$ whenever $n >N$ but this is clearly impossible if $\epsilon <1$. Hence $S$ is not fundamental in $\ell^{\infty}$.

There is no countable fundamental set in $\ell^{\infty}$ since this space is not separable.

The colection of al sequences $(a_n)$ such that $a_n=0$ for some $n$ is a fundamental set in $\ell^{\infty}$.