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, $y \in \Img g$.

$\blacksquare$