let's call a set $A \subseteq \mathbb{N}$ recursively enumerable if it's "partial characteristic function" $\tilde{\chi}_A$ is computable, whereby $\tilde{\chi}_A$ is defined as: $\tilde{\chi}_A$:= 1, if $x \in A$ and $\tilde{\chi}_A$ is undefined otherwise.
Now for $i = 0,1$ let $P_i$ be the set of all programmes p that terminate for input p and have output i. It's clear that $P_0 \cap P_1 = \emptyset$.
I'd be glad if you could help me with showing that the sets $P_0$ and $P_1$ are recursively enumerable, but that there's no decidable set E with $P_0 \subseteq E$, $P_1 \subseteq \mathbb{N}\backslash E$.
Looking forward to your suggestions!
The partial characteristic function of $P_0$ is computable by the algorithm: "On input $p$, run program $p$ with input $p$; if and when it terminates with output $0$, print $1$." $P_1$ is recursively enumerable similarly.
If they were separated by a decidable set $E$ as in the question, let $p$ be a program that outputs $1$ when $p\in E$ and outputs $0$ when $p\notin E$, i.e., $p$ computes the characteristic function of $E$. What would $p$ do when fed input $p$? It must eventually halt and output $0$ or $1$, since that's all $p$ ever does on any input. If it outputs $0$, then that means $p\in P_0\subseteq E$, so, by our choice of $p$, it should, on input $p$, compute $1$. Similarly, if $p$ on input $p$ outputs $1$, then $p\in P_1$, so $p\notin E$, and $p$ on input $p$ should output $0$. So we have a contradiction in either case. Therefore, no such program $p$ can exist.