I am stuck proving the following sets to be recursive or recursive enumerable (or none of the both).
first set: $$\{i\ |\ \exists n, \phi_i(n) \text{ converges and } \phi_i(n+1) \text{ converges}\}$$
second and third set:
$$\{i\ |\ D(\phi_i)\text{ disjoint }D(\phi_a)\}$$ in case $D(\phi_a)$ is not the empty set
$$\{i\ |\ D(\phi_i)\text{ disjoint }D(\phi_a)\}$$ in case $D(\phi_a)$ is the empty set
notation:
$\phi_i$ means the $i$-th computable function of $\phi$ and $D(p_i)$ denotes the set $\{x\ |\ \phi_i(x)\text{ converges}\}$
$\phi_a$ means the $a$-th computable function of $\phi$ and $D(p_a)$ denotes the set $\{x\ |\ \phi_a(x)\text{ converges}\}$
Thank you a lot for your help!!
The first set is r.e., but not recursive. It isn't recursive by Rice's theorem (one can also reduce from the halting problem by letting $\phi_{f(e)}(n) = \phi_e(e)$; then $f(e)$ is in your set if and only if $e \in H$). To see that it is r.e., dovetail all computations $\phi_i(n)$ together and enumerate $i$ whenever you see $\phi_i(n)\downarrow$ and $\phi_i(n+1)\downarrow$ for some $n$.
The second set is not r.e. (and not recursive). It is not recursive by Rice's theorem (one can also use the same reduction as above, and obtain that $f(e)$ is in your set if and only if $e \not\in H$). On the other hand, the set is co-r.e., as one can enumerate the complement by dovetailing all computations $\phi_i(n)$ together and enumerating $i$ if we see both $\phi_i(n)\downarrow$ and $\phi_a(n)\downarrow$ for some $n$. Since any function which is both r.e. and co-r.e. is recursive, the set cannot be r.e..
The third set is all of $\omega$ since every set is disjoint from the empty set, so it is recursive.