Why is $(e_n)$ not a basis for $\ell_\infty$?

Question

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.

Answer 1

There are basically two things to note here. First you need to understand what it means for a set to be a basis of an infinite dimensional normed space. In your context, I am all but certain that this means that it is a Schauder basis, which is a linearly independent set such that the set of all finite linear combinations of members of the set is dense in the space. The other kind of basis is a Hamel basis, which is an "algebraic" basis, i.e. a linearly independent set such that the set of all finite linear combinations of members spans the space.

For $\ell^\infty$, any member of the set of finite linear combinations of $e_n$ is eventually zero. But $[1,1,1,\dots] \in \ell^\infty$ but it is at least $1$ away from any sequence which is eventually zero.

Answer 2

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.