I have three related questions. The first two are extremely similar. The third one asks a question based on the first two. Let $K$ denote the diagonal halt set. Similarly let $bb:\mathbb{N} \rightarrow \mathbb{N}$ denote the busy beaver function and let $BB$ be the the set representing the image-set of the function $bb$.
(Q1) Define a set $A$ as follows:
if $x \notin BB$ then $x \notin A$
if $x \in BB$ then:
(a) if $x=bb(a)$ for some $a\in K$ then $x\in A$
(b) if $x=bb(a)$ for some $a\in K'$ then $x\notin A$
The question is simply whether $A$ is (i) co-r.e. OR (ii) neither r.e. nor co-r.e.?
(Q2) Define a set $B$ as follows:
if $x \notin BB$ then $x \notin B$
if $x \in BB$ then:
(a) if $x=bb(a)$ for some $a\in K$ then $x\notin B$
(b) if $x=bb(a)$ for some $a\in K'$ then $x\in B$
The question is simply whether $B$ is (i) co-r.e. OR (ii) neither r.e. nor co-r.e.?
(Q3) Does there exist a function $f:\mathbb{N} \rightarrow \mathbb{N}$ (denote the corresponding image-set as $F$) with the following properties: (i) $f$ is strictly increasing (ii) $f$ eventually grows faster any recursive function (iii) $F \equiv_T K$ (iv) $F$ is not co-r.e.
What I do know for sure is that if we replace $K$ with any recursive set the answer to both (Q1) and (Q2) is co-r.e. But because $K$ is not recursive, I do not know the answer. Though I guess(?) at least one of (Q1) and (Q2) has to be simpler than the other? (perhaps I should have given it more thought before posting the question)