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A total function $f: \mathbb{N} \to \mathbb{N}$ is monotone iff $f(x) < f(y)$ whenever $x < y$. Show that if $A$ is the range of a monotone recursive function then $A$ is recursive.

Hint: first show that the relations defined by $y = f(x)$ and $y < f(x)$ are recursive.

The hint part is easy. The graph relation of a recursive total function is a recursive relation. Since $f$ is a recursive total function, then the graph relation $y = f(x)$ is a recursive relation.

Using composition on the primitive recursive functions of modified (non-negative) difference and the signum function, we can see that $y < f(x)$ is recursive.

\begin{align*} \mathbf{1}_{y < f(x)}(x,y) &= \text{sg}(f(x) - y)) \\ \end{align*}

From there, if we can show a recursive construction of the following function then we will complete the problem:

\begin{align*} \mathbf{1}_{A}(y) &= \exists x (y = f(x)) \\ \end{align*}

This construction uses unbounded existential quantification, which is not necessarily recursive, so that won't solve the problem.

From here, I'm stuck on what to try.

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So the trick is to use the fact that it's monotone. For all monotone functions $f$ of the natural numbers you have that $n\leq f$. So to check whether $y$ is in the image all you need to do is check all $x\leq y$ since for any $x>y$ you have $f(x)>y$. This gives the bound you need.

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