= Solution
For $j\in\mathbb N_0$, let $e_j\in B$ have $e_j(j)=1$ and $e_j(k)=0$ for $k\ne j$. If $f,g\in F$ and $z\in\mathbb Z$, then $f+g$ can be nonzero only where $f$ or $g$ is nonzero, and $zf$ can be nonzero only where $f$ is nonzero. These are finite sets. Since $0\in F$, this proves that $F$ is a <submodule>.
Every $f\in F$ has the finite expression
$$
f=\sum_{j:\,f(j)\ne0}f(j)e_j.
$$
Evaluating at coordinate $k$ recovers the coefficient of $e_k$, so the expression is unique. Thus $\{e_j:j\in\mathbb N_0\}$ is a <basis of a module> for $F$, and
$$
\boxed{F\cong\bigoplus_{j\in\mathbb N_0}\mathbb Z}
$$
is a <free module>. By contrast, the ambient <module of integer sequences> $B=\prod_{j\in\mathbb N_0}\mathbb Z$ allows arbitrary integer sequences.
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