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

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.

• basis in what sense? May 7, 2015 at 19:58
• Because $(1, 1, 1, \dots , 1, \dots )$ is not spanned by those. May 7, 2015 at 19:59
• I suspect the meaning here is not a Hamel basis. May 7, 2015 at 20:00
• @Schauder basis May 7, 2015 at 20:01

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.

• That last line is nailing it. May 7, 2015 at 20:00
• @AsafKaragila In what manner? May 7, 2015 at 20:01
• People still wear pants? May 7, 2015 at 20:12
• @Na'omi I explained that above: a Schauder basis requires that the span of the finite linear combinations be dense. One reason why this requirement is essential is that in fact $\sum_{j=1}^\infty e_j$ is not a well defined sum in $\ell^\infty$ because $v_N=\sum_{j=1}^N e_j$ has no limit in $\ell^\infty$.
– Ian
Apr 14, 2019 at 23:34
• @Na'omi It will help to drop the notion that infinite sums are really any different from general limits. $\sum_{j=1}^\infty e_j$ is a meaningless symbol in the $\ell^\infty$ setting because $x_n=\sum_{j=1}^n e_j$ is not Cauchy in $\ell^\infty$. That is easy to see because $\| x_n-x_m \|_{\ell^\infty}=1$ for all $n \neq m$. That's all there is to it. The only question is whether this normed vector space type analysis is the right way to look at $\ell^\infty$ in the first place.
– Ian
Apr 15, 2019 at 14:31

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.

• I'm a little late with this answer but I thought I'd add my 5 cents anyway... hope it helps! May 8, 2015 at 12:11
• Technically a Schauder basis itself is not dense, but its rational span would be, while still being countable.
– Ian
Apr 17, 2019 at 12:42