If we generally would define the smallest set, that has to have some properties, as the set obtained by intersecting all the sets that have those properties (if the intersection is non-empty) and the tiniest set, that has to have some properties, as the set with the least cardinality, that has to have those properties (if there is only one set with the least cardinality), would then the set $F$ of primitive recursive function be also the tiniest set (besides being the smallest one) $F \subseteq \cup_{k\in \mathbb{N}} \left\{f:\mathbb{N}^k \rightarrow \mathbb{N} \right\}$, that contains the base functions and is closed under composition and primitive recursion ? (Or if I may rephrase the question: Is $F$ the only countable set, that contains the base functions and is closed under composition and primitive recursion ?)
Side question: I know $F$ is countable set, since to every function $f \in F$ there corresponds a very basic program containing only bounded loops and such; and a program is just a finite (but arbitrary long) string over a finite (and fixed) alphabet; thus $F$ can be seen as a reunion of all finite strings (that make synthactically sense; i.e. are programs) over a finite fixed alphabet - and therefore has to be countable. But how can I prove that $F$ is countable without falling back to interpret a function $f\in F$ as a program (just by staying "inside mathematics") ?