Arithmetic hierarchy I'm curious by Arithmetic Hierarchy. I'm trying to do the following ones:
1) How to show that $\{e\mid \mathrm{dom}(\phi_e)\text{ is recursive}\}$ is $\Sigma_3^0$?
For showing this, I need to really relate it to the characteristic function.
2) $F=\{e\mid \mathrm{dom}(\phi_e)\text{ is finite}\}$ is $\Sigma_2^0$?
Is finite a synonym for recursive?
Thanks
 A: Since it seems that you are interested in solving these problems yourself, I'll do a modified version of (2). You can adjust the proof to figure out (2), and that might give you some idea of how to attack (1). Let me just first clarify the notation I am using. $W_{e}$ is the domain of $\phi_{e}$. $T$ is the Kleene T-predicate. $S_{m}^{n}$ is Kleen's $S-m-n$ function, $\mu$ is the minimization operator, and $\# g$ is the code for a function $g$. 
Classify in the arithmetical hierarchy the set $A = \{e|W_{e} \mbox{ is finite and nonempty}\}$.
Solution: This set $A$ is given by the $\Sigma_{2}^{0}$ relation
$$ (\exists w \exists c \forall w' \forall c' \forall (n > 0))[T(e,w,c) \wedge (w + n= w' \to \neg T(e,w',c'))] $$
I.e. there exists an input $w$, and a code for computation on $w$, written here as $c$, such that for any other input $w'$ and any other code for the potential convergence of that input, not only does $e$ convergene on $w$ (as witnessed by $c$), but if $w'$ is larger than $w$ (so that there is some $n > 0$where $w+n = w'$), then any $c'$ does not witness convergence on $w'$.
Now, we prove that $A$ is $\Sigma_{2}^{0}$-complete. To see this, let $B$ be the $\Sigma_{2}^{0}$ set $\exists u \forall v Q(u,v,x)$, with $Q$ recursive. Define
$$ g(x,t) = \left\{ \begin{array}{cc}
0 & t = 0\\
\sum_{u \leq t} \mu_v \neg Q(u,v,x) & \mbox{otherwise}
\end{array} \right. $$
If $x \in B$, then for some $u$ and onward, this summation stops converging, as for that $u$ there is no $v$ such that $\neg Q(u,v,x)$ and so there is certainly no least such $v$! However, by construction, $g(x,t)$ always converges for $t= 0$, so the domain of $g$ is non-empty and finite. If $x \notin B$, then for every $u$, there is some $v$ such that $\neg Q(u,v,x)$, and so $g$ is total. Thus, the function
$$f :x \to S_{1}^{1}(\# g,x)$$
reduces $B$ to $A$.
