Just like in these two questions: 1, 2, I was also struggling to understand bases in $\ell^\infty$ in a constructive way, which I understood was not possible. However, in order to get a better understanding of infinite dimensional vector spaces and Hamel bases in general, I have a couple of perhaps simple questions:
Q1: Since any vector space has a Hamel basis, we still have a basis $B$ for $\ell^\infty$, even if we cannot write it down explicitly. This however means that for any vector $v\in\ell^\infty$, $v$ can be written uniquely as a linear combination of (finitely) many vectors $B \supset B_0=\{b_1,\dots,b_n\}$, which we would thus be able to construct given a specific $v_0\in \ell^\infty$ (right?) (EDIT:) not be able to construct, since we do not know $B$. So, knowing such a finite linear combination exist, regardless whether $B_0\subset B$ or not, are we at all able to construct a finite linear combination in general when not all but finitely many terms are $0$, and all of them are distinct, e.g. could we construct a finite $B_0$ be for say $\ell^\infty\ni v_0=(1/n)_{n=1}^\infty$?
Basically, what felt weird was:
Q1.2: Could we have a sequence in $\ell^\infty$ that thus can be written as a (finite) linear combination, but we may never be able to construct even one?
EDIT: But technically just the vector $v$ itself is a finite linear combination of vectors in $\ell^\infty$, so this question became rather trivial. I guess I was somehow confused by the counter intuitive notion of not being able to construct the basis. I think the answer given (together with this edit) suffice.
Q2: Is (more generally) a reason that $\{e_i\}$ does not form a (Hamel) basis for $\ell^\infty$, as mentioned in 1, essentially (but rather informally) that any assumed countable basis $\{b_j\}$ for any Banach space $X$ would have the property that any $B_n:=\text{span}\{b_1,\dots,b_n\}$ would of course be a proper subspace and thus a closed nowhere dense subset of $X$, and thus $X=\cup_n B_n$ would contradict Baire's (since $\ell^\infty$ is Banach)?
Best Answer
Q1: Of course we can't say what $B_0$ is without knowing what $B$ is! Given just that $B$ is some basis, $B_0$ could be any independent set that has $v_0$ in its span.
Q2: It's simply obvious that $(e_j)$ is not a Hamel basis for $\ell_\infty$; for example if $x=(1,1,1,\dots)$ then it's clear that $x$ is not a finite linear combination of the $(e_j)$.
In fact it's trivial to show directly that $(e_j)$ is not even a Schauder basis. Say $x = (1,1,1,\dots)$. For any $n$ and scalars $a_1,\dots,a_n$ we have $$||x-\sum_{j=1}^n a_je_j||\ge1.$$So $x\ne\sum_{j=1}^\infty a_j e_j$.