I'm trying to show that there is a pair of $\Sigma_1^0$ recursively inseparable sets.
From the definition, recursive inseparable is if there is no recursive set $C$ such that $A\subset C$ and $B\cap C = \emptyset$.
Also, my guess to the above is taking $\Delta_1^0$ with some complement. Thanks
Let $\{\Phi_e\}_{e \in \omega}$ be a effective listing of all partial computable functions. Let $A = \{e : \Phi_e(e) = 0\}$ and $B = \{e : \Phi_e(e) = 1\}$. Both are c.e. sets hence $\Sigma_1^0$. They are disjoint. The claim is that $A$ and $B$ form a computably inseparable pair. Suppose there exists a computable $C$ such that $A \subseteq C$ and $B \cap C = \emptyset$. Let $\chi_C$ denote the characteristic function of $C$. Since $C$ is computable, $\chi_C$ is a (total) computable function. There exists a $u$ such that $\chi_C = \Phi_u$. Since $\chi_C$ is total and a characteristic function (i.e. take value $0$ or $1$), either $u \in A$ or $u \in B$. Suppose $u \in A$, then $0 = \Phi_u(u) = \chi_C(u)$. $u \notin C$. Since $A \subset C$, $u \notin A$. Contradiction. Suppose $u \in B$. Then $1 = \Phi_u(u) = \chi_C(u)$. Then $u \in C$. However, $C \cap B = \emptyset$. Contradiction.