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Kreyszig1989 (quote unchanged inclusive exclamation mark):

More generally, if $X$ is any vector space, not necessarily finite dimensional, and $B$ is a linearly independent subset of $X$ which spans $X$, then $B$ is called a basis (or Hamel basis) for $X$. Hence if $B$ is a basis for $X$, then every nonzero $x\in X$ has a unique representation as a linear combination of (finitely many!) elements of $B$ with nonzero scalars as coefficients.

Now consider the vector space of sequences in $\mathbb{R}$ over $\mathbb{R}$, the basis $B=\{(\delta_{0k}),(\delta_{1k}),(\delta_{2k}),\ldots\}$, and the vector $x=(1,2,3,4,\ldots)$. Then there is no linear combination of finitely many elements of $B$ equating $x$.

What should I conclude? (i) $B$ is not a Hamel basis; (ii) the condition "finitely many" given by Kreyszig is too strong; (iii) something else.

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The condition given is spot on. In general, we only know how to add finitely many vectors, so it does not make sense to talk about linear combinations involving infinitely many nonzero vectors. So we do not do so. “Linear combination” in linear algebra always means “finitely many terms”.

Your set $B$ is linearly independent, but its span consists only of the almost null sequences: sequences that have only finitely many nonzero elements. As you note, your sequence $x$ is not in the span of $B$. So $B$ is not a (Hamel) basis for the entire space.

Assuming the Axiom of Choice, $B$ can be completed to a basis for your vector space.

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  • $\begingroup$ I guess that this is connected to the sentence "The preference of other types of bases for infinite-dimensional spaces is justified by the fact that the Hamel basis becomes "too big" in Banach spaces" in wikipedia $\endgroup$ Sep 24, 2019 at 6:14
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    $\begingroup$ @AaronLenz: Banach spaces have extra structure (a norm) and conditions (completion with respect to norms) that allows you to talk about convergence. So in a Banach space, one can work with “infinite sums” the same way we do in the real numbers with series: as limits of sequences of vectors. You are then going beyond just plain linear algebra and into the realm of analysis (aka “functional analysis” in this setting). It’s not that Hamel bases are “too big” in Banach spaces, bu that Banach spaces have other structure that lets you get away with “smaller” sets. $\endgroup$ Sep 24, 2019 at 6:21
  • $\begingroup$ thank you! it's only 8:24 and I have the feeling that it could be enough for today (and thanks to others) $\endgroup$ Sep 24, 2019 at 6:25
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You're right. $x$ is not a finite linear combination of the $\delta_{ik}$ ( I suppose you mean by those all vectors that are exactly $1$ at one coordinate and $0$ everywhere else) and so they do not span all the vectors and so are not a Hamel basis.

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$B$ is linearly independent but it does not span $X$. The span of $B$ consists precisely of sequences which have only finite number of non-zero coordinates. Hence it is not Hamel basis.

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  • $\begingroup$ why $B$ does not span $X$? because only finite linear combination are accepted? $\endgroup$ Sep 24, 2019 at 6:00
  • $\begingroup$ @AaronLenz: That’s not the correct way to phrase it; sets don’t span vectors. But yes, $x$ is not in the span of $B$. $\endgroup$ Sep 24, 2019 at 6:02
  • $\begingroup$ @ArturoMagidin I wrote $X$ meaning the space not the vector $x$ (as written by Kavi R. Murthy) $\endgroup$ Sep 24, 2019 at 6:12
  • $\begingroup$ @AaronLenz: So you did; my apologies. Yes: $B$ does not span the space of all real sequences. $\endgroup$ Sep 24, 2019 at 6:18

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