Set is Recursive iff Set and Complement are Recursively Enumerable

Theorem
Let $S \subseteq \N$ be a set of natural numbers.

Then:
 * $S$ is a recursive set


 * $S$ and $\N \setminus S$ are recursively enumerable sets.
 * $S$ and $\N \setminus S$ are recursively enumerable sets.

Necessary Condition
By Complement of Primitive Recursive Set, $\N \setminus S$ is a recursive set as well.

Thus, it suffices to show that every recursive $S$ is recursively enumerable.

By definition of recursive set, the characteristic function $\chi_S$ is recursive.

Define:
 * $\map f x = \map {\mu z} {\map {\overline {\operatorname{sgn}}} {\map {\chi_S} x}}$

As $f$ is obtained by substitution and minimization, and: it follows that $f$ is a recursive function.
 * Signum Complement Function is Primitive Recursive
 * Primitive Recursive Function is Total Recursive Function

We now show that $\Dom f = S$.

Suppose $x \in S$.

Then, by definition of characteristic function:
 * $\map {\chi_S} x = 1$

Thus, by definition of the signum-bar function:
 * $\map {\overline {\operatorname{sgn}}} {\map {\chi_S} x} = 0$

Therefore, by the definition of minimization:
 * $\map {\mu z} {\map {\overline {\operatorname{sgn}}} {\map {\chi_S} x}} = 0$

as the previous equation is true for every value of $z$, and $0$ is the smallest natural number.

Hence, $f$ is defined at $x$, so $x \in \Dom f$.

Now, suppose $x \notin S$.

By definition of characteristic function:
 * $\map {\chi_S} x = 0$

By definition of the signum-bar function:
 * $\map {\overline {\operatorname{sgn}}} {\map {\chi_S} x} = 1$

Therefore, by the definition of minimization:
 * $\map {\mu z} {\map {\overline {\operatorname{sgn}}} {\map {\chi_S} x}}$ is undefined

as there is no $z$ that causes the function to be equal to $0$.

Hence, $x \notin \Dom f$.

As we have shown $x \in S \iff x \in \Dom f$, it follows that $S = \Dom f$.

By Set is Recursively Enumerable iff Domain of Recursive Function, $S$ is then recursively enumerable.

Sufficient Condition
Let $S$ and $\N \setminus S$ be recursively enumerable.

Suppose $S$ is the empty set.

Then $\map {\chi_S} x = 0$ is the characteristic function of $S$.

In that case, $\chi_S$ is recursive by: and so $S$ is recursive by definition.
 * Constant Function is Primitive Recursive
 * Primitive Recursive Function is Total Recursive Function

Suppose instead that $\N \setminus S$ is the empty set.

Then $\map {\chi_S} x = 1$ is the characteristic function of $S$.

By the same argument, it follows that $S$ is recursive.

Now, suppose neither $S$ nor $\N \setminus S$ are empty.

Then, Recursively Enumerable Set is Image of Primitive Recursive Function/Corollary applies.

Therefore, there exist primitive recursive $f, g : \N \to \N$ such that:
 * $S = \Img f$
 * $\N \setminus S = \Img g$

Define:
 * $\map h x = \map {\mu z} {x = \map f z \lor x = \map g z}$

By: $h$ is a recursive function.
 * Equality Relation is Primitive Recursive
 * Set Operations on Primitive Recursive Relations
 * Minimization on Relation Equivalent to Minimization on Function

Finally, define:
 * $\map {\chi_S} x = \map {\chi_{\operatorname{eq}}} {x, \map f {\map h x}}$

$\chi_S$ is recursive, as it is defined by substitution on recursive functions.

We now show that $\chi_S$ is the characteristic function of $S$.

Suppose $x \in S$.

Then, $x \in \Img f$ and $x \notin \Img g$.

That is:
 * There is some $z \in \N$ such that $\map f z = x$
 * There is no $z \in \N$ such that $\map g z = x$

Therefore, $\map h x$ is defined to be the least $z$ such that $\map f z = x$.

Thus, $\map {\chi_S} x = 1$, as expected.

Suppose $x \notin S$.

Then, conversely, $x \notin \Img f$ and $x \notin \Img g$.

The same argument applies, and:
 * There is no $z \in \N$ such that $\map f z = x$
 * $\map h x$ is the least $z$ such that $\map g z = x$, which necessarily exists

Thus, $\map f {\map h z} \ne x$, and $\map {\chi_S} x = 0$.

Since the characteristic function of $S$ is recursive, $S$ is a recursive set by definition.