Recursively Enumerable Set is Image of Primitive Recursive Function
Theorem
Let $S \subseteq \N$ be recursively enumerable.
Suppose $S$ is non-empty.
Then there exists a primitive recursive function $f : \N^k \to \N$ such that:
- $\Img f = S$
Corollary
Let $S \subseteq \N$ be recursively enumerable.
Suppose $S$ is non-empty.
Then there exists a primitive recursive function $f : \N \to \N$ such that:
- $\Img f = S$
Proof
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By definition of recursively enumerable, there exists a recursive function $g : \N^\ell \to \N$ such that:
- $\Img g = S$
By definition of non-empty, there exists some $x \in S$.
By Kleene's Normal Form Theorem, there exist:
- A primitive recursive relation $T \subset \N^{\ell + 2}$
- A primitive recursive function $U : \N \to \N$
such that a partial function $h : \N^\ell \to \N$ is recursive if and only if there exists $e \in \N$ such that:
- $\map h {x_1, \dotsc, x_\ell} \approx \map U {\mu z \map T {e, x_1, \dotsc, x_\ell, z}}$
for every $\tuple {x_1, \dotsc, x_\ell} \in \N^\ell$.
Thus, as $g$ is recursive, there exists such an $e$.
Define $V : \N^{\ell + 1} \to \N$ as:
- $\map V {x_1, \dotsc, x_\ell, t} = \map {\mu z \le t} {\map T {e, x_1, \dotsc, x_\ell, z}}$
$V$ is primitive recursive by Bounded Minimization is Primitive Recursive.
Define $f : \N^{\ell + 1} \to \N$ as:
- $\map f {x_1, \dotsc, x_\ell, t} = \begin{cases}
\map U {\map V {x_1, \dotsc, x_\ell, t}} & : \map V {x_1, \dotsc, x_\ell, t} \le t \\ x & : \map V {x_1, \dotsc, x_\ell, t} > t \end{cases}$
By Definition by Cases is Primitive Recursive and Ordering Relations are Primitive Recursive, $f$ is primitive recursive.
It remains to show that $\Img f = \Img g$.
Suppose $y \in \Img g$.
Then, for some $\tuple {x_1, \dotsc, x_\ell} \in \N^\ell$:
- $\map g {x_1, \dotsc, x_\ell} = y$
Thus, by definition of $T$ and $U$:
- $\map U {\map {\mu z} {\map T {e, x_1, \dotsc, x_\ell, z}}} = y$
Let $t = \map {\mu z} {\map T {e, x_1, \dotsc, x_\ell, z}}$.
Hence:
- $\map V {x_1, \dotsc, x_\ell, t} = \map {\mu z} {\map T {e, x_1, \dotsc, x_\ell, z}} = t$.
Therefore, as $t \le t$:
- $\map f {x_1, \dotsc, x_\ell, t} = \map U t = y$.
Thus, $y \in \Img f$.
$\Box$
Now, suppose $y \in \Img f$.
Then there exists $\tuple {x_1, \dotsc, x_\ell, t} \in \N^{\ell + 1}$ such that:
- $\map f {x_1, \dotsc, x_\ell, t} = y$
By the definition of $f$, either:
- $\map U {\map V {x_1, \dotsc, x_\ell, t}} = y$ and $\map V {x_1, \dotsc, x_\ell, t} \le t$, or
- $x = y$ and $\map V {x_1, \dotsc, x_\ell, t} > t$.
In the second case, $x \in S = \Img g$ by definition.
In the first, let $r = \map V {x_1, \dotsc, x_\ell, t}$.
As $r \le t$, by definition of $V$ and bounded minimization:
- $r = \map {\mu z} {\map T {e, x_1, \dotsc, x_\ell, z}}$
But then, $y = \map U r = \map U {\map {\mu z} {\map T {e, x_1, \dotsc, x_\ell, z}}} = \map g {x_1, \dotsc, x_\ell}$.
Therefore, $y \in \Img g$.
$\blacksquare$