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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.

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    $\begingroup$ basis in what sense? $\endgroup$
    – Studzinski
    May 7, 2015 at 19:58
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    $\begingroup$ Because $(1, 1, 1, \dots , 1, \dots )$ is not spanned by those. $\endgroup$
    – Crostul
    May 7, 2015 at 19:59
  • $\begingroup$ I suspect the meaning here is not a Hamel basis. $\endgroup$
    – Asaf Karagila
    May 7, 2015 at 20:00
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    $\begingroup$ @Schauder basis $\endgroup$
    – user62498
    May 7, 2015 at 20:01

2 Answers 2

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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.

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  • $\begingroup$ That last line is nailing it. $\endgroup$
    – Asaf Karagila
    May 7, 2015 at 20:00
  • $\begingroup$ @AsafKaragila In what manner? $\endgroup$ May 7, 2015 at 20:01
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    $\begingroup$ People still wear pants? $\endgroup$ May 7, 2015 at 20:12
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    $\begingroup$ @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 $\ell^\infty$. $\endgroup$
    – Ian
    Apr 14, 2019 at 23:34
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    $\begingroup$ @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. $\endgroup$
    – Ian
    Apr 15, 2019 at 14:31
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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.

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

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