I think that the set of recursive subsets of the set of natural numbers is not r.e. (meaning the set of codes of those subsets according to an admissible numbering).

I think I can provide the sketch of the proof of such statement and I would like to ask if what I write is correct.

**PROOF SKETCH**

This statement is equivalent to prove that the set of characteristic functions of recursive sets of natural numbers is not r.e., we are talking about recursive functions of this form: $f: mathbb{N} longrightarrow {0,1}$.

To be precise we are talking of the set of the codes according to an admissible numbering of those functions.

For any such function $f_k$ there is an infinite recursive binary sequence: $f_k(0), f_k(1),…,f_k(i),…: i in mathbb{N}$.

Let $TM_k$ be a Turing machine that on empty input outputs the binary sequence associated to $f_k$.

Let $widetilde{TM}$ be a Turing machine that enumerates those Turing machines $TM_k: k in mathbb{N}$ (to be precise it enumerates the codes of those Turing machines).

We can build a Turing machine $widehat{TM}$ that given as input the code of any Turing machine $TM_x$ runs both $TM_x$ and $widetilde{TM}$.

Then one of those two things must happen, either $TM_x$ halts or $widetilde{TM}$ outputs the code of

$TM_x$ which means $TM_x$ does not halt. So $widetilde{TM}$ can decide the halting problem (restricted to $TM$‘s with empty input but it is equivalent to the general halting problem, see this link), which is a contradiction.

So $widetilde{TM}$ can not be built, the set of $TM_k$ is not r.e. and so the set of recursive subsets of the set of natural numbers is not r.e. $blacksquare$

I assume the Turing degree of this set is $mathbf{0}’$ because an oracle for the halting problem make this set recursive.