Proving that a separating set is not computable Let $X =$ {$d$ : $\psi_d(0)=0\}$ and $Y$ = { $d$ : $\psi_d(0) = 1$ }. Show that if $S\subseteq \mathbb{N}$ has the property $X \subseteq S$ and $Y \cap S = \emptyset$ then $S$ is not recursive.
So I was thinking if I could find some function $f$ such that $\psi_{f(d)}(x) = \psi_d(d)$ for all $d$ and $x$.  And then we were to consider the characteristic function of $\{t: f(t) \in S\}$ which we could call $\psi_a$ and see that computing $\psi_a(a)$ can be paradoxical. But how do we show we can find such a function $f$ and does my thinking sound right?
 A: I think that the ideas in the post are completely on track. However, when I tried to write up the answer as a hint, I kept making mistakes. In the end, it seemed that the recursion theorem was useful to solve the problem. Otherwise, I was trying to re-prove it in my solution.  
For the diagonalization, we want a $d$ so that $d \in X$ if and only if $d \not \in S$.   Expanding that, we want $\psi_d(0) = 0$ if and only if $\chi_S(d) = 0$.  Thinking intuitively, we want $\psi_d(0)$ to simply compute $\chi_S(d)$, which we can achieve with the recursion theorem. 
Assuming $S$ is computable, we have $\chi_S = \psi_e$ for some $e$. Consider the function $g$ so that $\psi_{g(z)}(0) = \psi_e(z) = \chi_S(z)$. We know $g$ is total computable from the s-m-n theorem. Therefore, by the recursion theorem, $g$ has a fixed point $d$ so that $\psi_d(0) = \psi_{g(d)}(0)$. 
Then we have 
$$d \in X \leftrightarrow \psi_d(0) = 0 \leftrightarrow \psi_{g(d)}(0) = 0 \leftrightarrow \chi_S(d) = 0 \leftrightarrow d \not \in S.$$
