Set is Recursively Enumerable iff Domain of Recursive Function

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

Then:
 * $S$ is recursively enumerable


 * There exists a recursive function $f : \N \to \N$ such that $S = \Dom f$
 * There exists a recursive function $f : \N \to \N$ such that $S = \Dom f$

where $\Dom f$ is the domain of $f$.

Necessary Condition
Suppose $S$ is the empty set.

Define $\map f x = \map {\mu z} {z \ne z}$.

Then, $f$ is undefined for every $x \in \N$, and thus is defined on the empty set.

Additionally, $f$ is recursive by:
 * Equality Relation is Primitive Recursive
 * Set Operations on Primitive Recursive Relations
 * Primitive Recursive Function is Total Recursive Function

Otherwise, let $S$ be non-empty.

Then, by Recursively Enumerable Set is Image of Primitive Recursive Function/Corollary, there exists a primitive recursive function $g : \N \to \N$ such that:
 * $S = \Img g$

Define $\map f x = \map {\mu z} {\map g z = x}$.

It remains to be shown that $\Dom f = \Img f$.

Let $y \in \Dom f$.

Then, there exists $z \in \N$ such that $\map g z = y$.

But then, $y \in \Img g$.

Now, let $y \in \Img g$.

Then, there exists $z \in \N$ such that $\map g z = y$.

Let $z'$ be the smallest such $z$.

Then, $\map f y = z'$ be definition.

Thus, $y \in \Dom f$.

Sufficient Condition
Define $\map g x = \map {\pr_1^2} {x, \map f x}$.

As projection is a basic primitive recursive function, and $g$ is obtained by substitution, it is recursive by definition.

It remains to be shown that $\Img g = \Dom f$.

Let $y \in \Img g$.

Then, by definition of substitution, $\map f y$ is defined.

Therefore, $y \in \Dom f$.

Now, let $y \in \Dom f$.

Then, as $\map f y$ is defined:
 * $\map g y = y$

Thus, $y \in \Img g$.