Is there a non-computable set $X$ and a a partial computable function $F$ such that for all natural numbers $e$, if $\Phi_e^X(0)$ halts, then $$F(e) > \Phi^X_e(0)?$$
If $X$ computable, then there is such an $F$. For example, $F$ might have the instructions
- Run $\Phi_e(0)$
- Add 1 to the result
- Halt.
However, if $X \geq_T 0'$, then there is no such $F$. To see this, suppose there were. Then, by s-m-n theorem, the following function $G$ would be $0'$-computable.
$$\Phi^X_{G(e)}(x) = \begin{cases} F(e) + 1 &\text{if } \Phi_e^X(0) \text{ converges}\\ 1 &\text{if } \Phi_e^X(0)\text{ diverges}. \end{cases} $$
Then, by the recursion theorem, there is $a$ such that $\Phi^X_{G(a)} = \Phi^X_a$. The computation $\Phi^X_a(0)$ cannot diverge as if so, then $\Phi^X_a(0) =\Phi^X_{G(a)}(0) = 1$, a contradiction. Therefore, $F(e) = \Phi_a^X(0) = \Phi_{G(a)}^X(0) < F(e)$, a contradiction.