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.
Best Answer
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.