= Solution
{c}
Every $v\in V$ is a <finitely supported sequence>, so for a finite <set> of indices $S$,
$$
v=\sum_{i\in S}v_i e^{(i)}.
$$
Thus $E$ is a <spanning set>. If $F$ contained $v\notin E$, the displayed expression would give a nontrivial relation
$$
v-\sum_{i\in S}v_i e^{(i)}=0
$$
among the distinct <vectors> of the finite <subset> $\{v\}\cup\{e^{(i)}:i\in S\}$ of $F$: the <coefficient> of $v$ is $1$. This contradicts the <linear independence> of $F$. It also covers $v=0$, whose singleton is not <linearly independent>. Hence \b[$F=E$]. In particular, $E$ is both a <basis> and a maximal <linearly independent> <subset> of $V$.
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