I'd appreciate some help for the following exercise. $\Sigma_1, \Pi_1$ and $\Delta_1$ are defined here: https://en.wikipedia.org/wiki/Arithmetical_hierarchy I think I can prove 2. when I have 1. - can someone give me hint, how to start with the proof? I already know that all $\Sigma_1$-sets are r.e. and all $\Pi_1$-sets are co-r.e. and all $\Delta_1$-sets are recursive.
Assume $S_0, S_1 \subset \mathbb{N}$ are $\Sigma_1$-sets which satisfy $S_0 \cup S_1 = \mathbb{N}$. Prove that there is a $\Delta_1$-set $E$ such that $S_0 \setminus S_1 \subset E \subset S_0$ is satisfied.
Assume $Q_0, Q_1 \subset \mathbb{N}$ are $\Pi_1$-sets which satisfy $S_0 \cap S_1 = \emptyset$. Prove that there is a $\Delta_1$-set $E$ such that $Q_0 \subset E$ and $Q_1 \subset \mathbb{N} \setminus E$ are satisfied.
EDIT:
I'm trying to construct $\Pi_1$ and $\Sigma_1$ formulas for $E$:
$\exists k \varphi_0(k,n),\exists k \varphi_1(k,n)$ are $\Sigma_1$ formulas which define $S_0$ and $S_1$.
$S_0=\{n\in \mathbb{N}: \mathbb{N}\vDash \exists k ~ \varphi_0(k,n)\}$
$S_1=\{n\in \mathbb{N}: \mathbb{N}\vDash \exists k ~ \varphi_1(k,n)\}$
$S_0\setminus S_1=\{n\in \mathbb{N}: \mathbb{N}\vDash \exists k ~ \varphi_0(k,n) \land \forall l ~ \lnot \varphi_1(l,n)\}$
$E=\{n\in \mathbb{N}: \mathbb{N}\vDash \exists k ~ \varphi_0(k,n) \land \exists l ~ \lnot \varphi_1(l,n)\}$
$\exists k ~ \varphi_0(k,n) \land \exists l ~ \lnot \varphi_1(l,n)$ is equivalent to a $\Sigma_1$ formula. Now I need a $\Sigma_1$ formula for $\mathbb{N} \setminus E$ or a $\Pi_1$ formula for $E$. Can someone give me a hint how to find it?
Here is the intuition: We know any natural number is in $S_0$ or in $S_1.$ So given $n\in\mathbb N,$ recursively enumerate $S_0$ and $S_1$ side by side. If $n$ appears first in the $S_1$ enumeration, $n\notin E,$ whereas if it appears first in the $S_0$ enumeration, $n\in E.$ (And break ties however.)