TITLE: Not a basis for $ l^\infty$ then what is it?
QUESTION [6 upvotes]: We know that $ l^\infty$ has not a Schauder basis and its Hamel basis is uncountably infinite.
Let $e_n=(e_{n1}, e_{n2},...)$ (for each $n\in \mathbb{N}$) s.t. $e_{nj}=0$ when $n\neq j$ and $e_{nn}=1$, now for $x=(x_1,x_2,\ldots)\in l^\infty$ we have:
$$x=\sum_{n=1}^\infty x_ne_n$$
$\{ e_n \}_{n\in\mathbb{N}}$ is not a Hamel basis and a Schauder basis then what is it?

REPLY [2 votes]: It is not true that $$x=\sum_{n=1}^{\infty} x_ne_n$$ for $x\in\ell^{\infty}$, because convergence does not necessarily occur in the $\ell^{\infty}$-norm. Indeed, if  $x=(1,1,1,\ldots)$, then, for any $N\in\mathbb N$, $$\left\|x-\sum_{n=1}^{N}x_ne_n\right\|_{\infty}=\|(\underbrace{0,\ldots,0}_{N\text{ times}},1,1,1,\ldots)\|_{\infty}=1,$$ so $x$ cannot be the norm limit of the partial sums $\{\sum_{n=1}^{N}x_ne_n\}_{N\in\mathbb N}$. Therefore,  $\{e_n\}_{n\in\mathbb N}$ is not a Schauder basis.
But even if it were a Schauder basis, it would still not be a Hamel basis, because this would require that any element $x\in\ell^{\infty}$ must be a finite linear combination of elements of $\{e_n\}_{n\in\mathbb N}$, not merely a norm limit of the partial sums.