Complement of Primitive Recursive Set

Theorem
Let $S \subseteq \N$ be primitive recursive.

Then its relative complement $\N \setminus S$ of $S$ in $\N$ is primitive recursive.

Proof
By definition, we have that the characteristic function $\chi_{\N \setminus S} \left({n}\right) = 1$ iff $\chi_S \left({n}\right) = 0$.

So $\chi_{\N \setminus S} \left({n}\right) = \chi_{\left\{{0}\right\}} \left({\chi_S \left({n}\right)}\right)$.

Thus $\chi_{\N \setminus S}$ is obtained by substitution from $\chi_{\left\{{0}\right\}}$ and $\chi_S$.

The result follows from Set Containing Only Zero is Primitive Recursive.