Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 2 / 4H / b

Fix an effective numbering of Turing machines and use the diagonal halting setIt is computably enumerable: dovetail all the computations and output each index when its computation halts. It is not a recursive set. Otherwise a machine could, on input $e$, loop if the alleged decider says $e\in H$ and halt otherwise. Applying this machine to its own index contradicts either answer. Hence $H$ is computably enumerable but not a recursive set.

With the diagonal halting set $H$, defineIf $A$ were computably enumerable, its odd-index inverse image would make $\mathbb{N}\setminus H$ computably enumerable. If $\mathbb{N}\setminus A$ were computably enumerable, its even-index inverse image would do the same. But a set and its complement being computably enumerable would make the set a recursive set: run the two enumerations in parallel until the input appears in one. This contradicts the undecidability of $H$. Thus neither $A$ nor its complement is computably enumerable.