Set of Even Numbers is Primitive Recursive

Theorem
Let $$E \subseteq \N$$ be the set of all even natural numbers.

Then $$E$$ is primitive recursive.

Proof
If $$n \in E$$ then $$n$$ is of the form $$n = 2 k$$ where $$k \in \N$$.

We have that:
 * if the characteristic function $$\chi_E \left({n}\right) = 1$$ then $$\chi_E \left({n + 1}\right) = 0$$.
 * if the characteristic function $$\chi_E \left({n}\right) = 0$$ then $$\chi_E \left({n + 1}\right) = 1$$.

So $$\chi_E$$ can be defined by:
 * $$\chi_E \left({n}\right) = \begin{cases}

1 & : n = 0 \\ \overline{\sgn} \left({\chi_E \left({n-1}\right)}\right) & : n > 0 \end{cases}$$

So $$\chi_E$$ is obtained by primitive recursion from the constant $1$ and the primitive recursive function $\overline{\sgn}$.

Hence the result.