TITLE: Why is $(e_n)$ not a basis for $\ell_\infty$?
QUESTION [5 upvotes]: Let $(e_n)$ (where $ e_n $ has a 1 in the $n$-th place and zeros otherwise) be unit  standard vectors of $\ell_\infty$. 
Why is $(e_n)$ not a basis for $\ell_\infty$?
Thanks.

REPLY [4 votes]: A Schauder basis is a dense countable subset. But there does not exist any dense countable subset in $\ell^\infty$.
Let us prove that no such set can exist:
By contradiction, assume that $D$ was a dense countable subset of $\ell^\infty$. Now consider the set $S = \{s\mid s_n \in \{0,1\}\}$, that is, all sequences only consisting of $0$ and $1$.
If we consider balls of radius ${1\over 3}$ around each $s$ then in every such ball there will have to be a $d$ in $D$ since $D$ is dense. But $S$ is uncountable therefore this cannot be the case. Hence $\ell^\infty$ cannot contain a dense countable subset.
In particular, $e_n$ cannot be a Schauder basis.