So basically I am trying to prove that a set $A = \{ x\in \mathbb{N} | \phi_x(n) \leq 2021, \forall n \}$ is not recursively commutable.
To do so, I wrote down $A$ as an index set $$ A = I_C = \{ x| \phi_x \in C\} \, \text{ where } C = \{ \phi : \mathbb{N} \rightarrow \mathbb{N} | \phi \downarrow \wedge \phi(n)\leq 2021, \forall n \} $$ Which allows me to say, based on Rice theorem, that the set is non recursive, but that's not enough.
So I want to use one of the things we proved during the lessons and prove, that an undefined function belongs to the set $I_C$, where an undefined function is meant to be a function, that runs in a cycle forever for any input provided.
I do know, that such function, let us call it $f$ is commutable (I don't know the proof though, nor have I found one on the internet).
What I don't understand is how do I go about proving that the function $f$ satisfies the second condition of returning a value $\leq 2021$, when it is not supposed to return a value at all. Or maybe I don't even need to prove it.
What I want to do is to prove, that $f \in C$, which will allow me to say that $\bar{K} \leq_m I_C$, which would ultimately mean that $I_C$ is non recursively enumerable, since $\bar{K}$ is not.
Suppose that $A$ were recursively enumerable. Define $h(x) = 0$, and define $g$ such that $\phi_{g(x)} = h \circ \phi_x$. Then $g$ is recursive, $\phi_{g(x)}(n) \downarrow$ iff $\phi_x(n) \downarrow$, and whenever $\phi_{g(x)}(n) \downarrow$, we have $\phi_{g(x)}(n) = 0 \leq 2021$.
Then consider the set $B = \{x \in \mathbb{N} | \forall n \in \mathbb{N}, \phi_x(n) \downarrow\}$. It is well-known that $B$ is not recursively enumerable.
But just in case you are not aware of the proof, suppose $B$ were recursively enumerable. Then since $B$ is infinite, there would be a computable surjection $f : \mathbb{N} \to B$. Then define $g : \mathbb{N} \to \mathbb{N}$ by $g(n) = \phi_{f(n)}(n) + 1$. Then $g$ is computable since $\phi$ and $f$ are computable, and $g$ is total since for all $n$, $\phi_{f(n)}$ is total.
Then take $m$ such that $g = \phi_m$. Since $g$ is total, take $n$ such that $f(n) = m$. Then $g(n) = \phi_{f(n)}(n) + 1 = \phi_{f(n)}(n)$. Contradiction.
Back to the proof. Since $A$ is recursively enumerable, there is a recursive function $w$ such that for all $n$, $w(n) \downarrow$ iff $w \in A$. Then we see that $x \in B$ iff $g(x) \in A$ iff $w(g(x)) \downarrow$. That is, for the computable function $z = w \circ g$, we have $x \in B$ iff $z(x) \downarrow$. Then $B$ is recursively enumerable. Contradiction.