Consider the set $S = \{x : \phi^1_x(x) \ \ \text{is undefined/does not converge\} }$
This is supposed to be a set that is not recursively enumerable. How do we prove this?
My thoughts so far:
Consider the sets $T_i$ that are the domains of the functions $\phi^1_i$.
The complement, $\bar S = \{x : \phi^1_x(x) \ \ \text{is defined} \} = \{x : x \in T_x\}$ (how legit is this line)??
The complement of the complement of $S$ is then $\{x : x \not\in T_x\}$ which is not r.e. because it is not the any of the sets $T_i$
Is this OK?
Is there a better argument?
As Rob Arthan says in a comment, what you've got looks like it's on the right track.
Here's something along similar lines.
Suppose that $S$ is r.e. Since every r.e. set is the domain of a partial recursive function, it follows that there's a $k$ such that
$\phi_k(x)$ converges iff $x\in S$, for all $x$.
Hence,
$\phi_k(k)$ converges iff $k\in S$,
but the choice of $S$ implies that
$k\in S$ iff $\phi_k(k)$ doesn't converge,
a contradiction.
You may recognize the set you mention in this article...
https://en.wikipedia.org/wiki/Creative_and_productive_sets