Suppose that f ∈ℱ1 is a recursive function and assume that...

Suppose that $f \in \mathcal{F}_{1}$ is a recursive function and assume that its image is infinite. Show that there exists a recursive function $g \in \mathcal{F}_{1}$ that is recursive and injective and satisfies $\operatorname{lm}(f)=\operatorname{lm}(g)$. Conclude from this that there exists an injective recursive function whose image is not recursive.

Key Concepts

Recursive Functions

A recursive function is one that can be computed by an algorithm; it is a function for which there exists a Turing machine (or equivalent computational model) that will produce its output in a finite amount of time for every input. These functions form the backbone of computability theory, ensuring that they are effectively calculable.

Injective Functions

An injective function (or one?to?one function) assigns distinct outputs to distinct inputs. In computability contexts, constructing an injective recursive function allows us to ‘reorder’ or ‘filter’ the outputs of another function while preserving its overall set of values, which is essential in many reductions and diagonalization arguments.

Image (Range) of a Function

The image of a function, also called its range, is the set of all outputs it produces. In the context of computability, even though a function is recursive (and hence its values can be computed), its image might still be 'complicated' (non-recursive) if it fails to be decidable, which reveals subtle distinctions between different levels of computability.

Recursive vs Non?recursive Sets

A set is recursive (decidable) if there is an algorithm that can determine in finite time whether any given element belongs to that set, while a non?recursive set lacks such a decision procedure. The result that there exists an injective recursive function whose image is not recursive emphasizes that even though a function may be simply computable, the set of its outputs may not be decidable, highlighting an important nuance in recursion theory.

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