How to argue that a set is recursive or recursively enumerable? I have the two sets listed below, and I want to argue whether each of them is recursive, recursively enumerable or neither recursive nor recursively enumerable.


*

*the set $A = \{ i | \text{Dom}(\phi_i) = \emptyset \}$, where $\text{Dom}(\phi_k) = \{x | \phi_k(x) \downarrow \}$, and where $\phi_k(x) \downarrow$ means that function with index $k$ converges at input $x$.

*the set $B$ of all total recursive functions $a : \mathbb{N} \rightarrow \mathbb{N}$ such that $a(n+1) \geq a(n)$, for all $n \in \mathbb{N}$
Please note that I have this definition for recursive sets, and this definition for recursive enumerable sets (r.e.). I also know that, if a set $A$ is recursive, then $A$ is also r.e.
The problem is that, I don't know how to argue about whether the sets above are recursive, r.e., or none of them. Any ideas about the problems?
EDIT:
$$\text{Halt}(x,y) = \begin{cases} 1, & \mbox{if } \phi_x(y) \downarrow \\ 0, & \mbox{if } \phi_x(y) \uparrow \end{cases}$$
where $\phi_x(y) \downarrow$, means function with index $x$ converges on input $y$, and $\phi_x(y) \uparrow$ means that it diverges.
 A: Let $K$ be the halting problem set $\{x \in \mathbb{N} \mid \phi_x(x) \downarrow\}$. It is known that $K$ is not recursive and hence (by Post's theorem) its complement $\overline{K}$ is not r.e. We say that $X$ is m-reducible to $Y$ (notation: $X \leq_\text{m} Y$) if there is a total recursive function $f(x)$ with $x \in X \Leftrightarrow f(x) \in Y$. It is quite easy to show that if $X \leq_\text{m} Y$ and $Y$ is recursive (r.e.), so is $X$.
To show that $A$ is not r.e. (and hence is not recursive) we will show that $\overline{K} \leq_\text{m} A$.
Consider the following function
$$f(x, y) = \begin{cases}
1,& x \in K\\ 
\uparrow,& x \notin K
\end{cases}$$
It is partial recursive and by the $s_m^n$-theorem there is a total recursive $g(x)$ with $\phi_{g(x)}(y) = f(x, y)$. It holds that $$x \notin K \Leftrightarrow (\forall y) \phi_{g(x)}(y) = f(x, y) =\,\uparrow\, \Leftrightarrow g(x) \in A.$$
Now assume that $B$ is r.e. and let $U(n, x) = a_n(x)$ be a total recursive universal function for this set, that is, the functions from $B$ form a sequence $U(0, x), U(1, x), \dots$. Consider the following function:
$$g(n) = \sum_{i = 0}^n U(i, n) + 1.$$
It is clearly a total recursive function. Since $U(i, n+1) \geq U(i, n)$ it holds that $$g(n+1) = \sum_{i = 0}^n U(i, n+1) + U(n+1, n+1) + 1 \geq \sum_{i = 0}^n U(i, n) + 1 = g(n),$$
hence $g(n) \in B$. On the other hand, if for some $m$ we have $U(m, n) = g(n)$, then taking $n = m$ we obtain $$U(m, m) = \sum_{i=0}^m U(i, m) + 1,$$ which is false. It means that $g(n) \notin B$, a contradiction. Hence $B$ is not r.e. (and, of course, is not recursive).
