Is it true that the domain of this function is finite? (This is a subproblem that came up as I was proving that $\{e: \phi_e(x) \text{ has infinite domain}\}$ is $\Pi_2$-complete.)
Let $A$ be a set of natural numbers and suppose $$n\in A\iff \forall y_1\exists y_2 T(n,y_1,y_2)$$
(where $T$ is a (computable) relation).
Consider this  partial function on pairs of natural numbers:
$$(n,x) \mapsto 1\text{ if }\forall y_1\leq x\exists y_2 T(n,y_1,y_2)\\(n,x)\text{ is undefined otherwise} $$
If $n\in A$, then the above function as a function of $x$ is total. Is it true that if $n\notin A$, then this function (again as a function of $x$) has finite domain? I think so, and I think this is a trivial result but I can't wrap my head around proving it in a formal way.
Here's what I think informally: if $n\notin A$, then the first condition in the definition may be true for small $x$, but starting from some $x_0$, that wouldn't be true. Is this indeed the case? I can't connect this with the negation of $x\in A$ i.e. with $\exists y_1\forall y_2 \neg  T(n,y_2,y_3) $.

Edit: I modified the original function. Does the new function has a finite domain as a function of $x$ when $x\notin A$? If not, I was wondering if there's a way to modify this function further so that (1) it remains computable, (2) if $n\in A$, the domain is still infinite, (3) if $n\notin A$, the domain is finite?
 A: I don't know why you insist that $(n,x)$ not be defined when it doesn't take the value $1$. I prefer to think of it as a total function that takes the value $0$ when $\forall y_1 \le x, \exists y_2 T(n,y_1,y_2)$ is not true, i.e. when $\exists y_1 \le x, \forall y_2, \neg T(n,y_1,y_2)$. Then the question is whether the set of $x$ for which $(n,x)$ maps to $1$ is finite or not.
Since the above function is now defined everywhere, let me call it $F(n,x)$. This function satisfies the following property: if $F(n,x) = 1$ and $x' \le x$, then $F(n,x') = 1$ as well. So there are two possibilities:

*

*Either $F(n,x) = 1$ for infinitely many integers, in which case the existence of the increasing sequence of positive integers $\{x_m\}_{m \in \mathbb N}$ such that $F(n,x_m) = 1$ implies that $F(n,x) = 1$ for all $x \in \mathbb N$. When this is the case, $n \in A$.


*Or, $F(n,x) = 1$ for finitely many integers. By the same argument as above, this means that if we denote the largest such integer by $x_0$, then $F(n,1) = F(n,2) = \cdots = F(n,x_0) = 1$, and $F(n,x) = 0$ for $x > x_0$ because $x=x_0 + 1$ would be the first integer for which $\forall y_2, \neg T(n,x_0+1, y_2)$. When this is the case, $n \notin A$.
Hope that helps,
A: Let $V$ denote the partial function from the question. Suppose $n\notin A$. Then $\exists x_0 \forall y_2 \neg T(n, x_0, y_2)$. Then the domain of $V(n,-)$ is a subset of $\{x\in N: x < x_0 \}$ (hence finite). Here's why nothing outside this finite set can lie in the domain. If $x \ge x_0$, then the condition $$\forall y_1 < x \exists y_2 T(n,y_1,y_2)$$
or equivalently $$\forall y_1\exists y_2 (\neg (y_1 \le x) \lor T(n,y_1,y_2))$$
is violated. Indeed, its negation is $$\exists y_1\forall y_2 ((y_1 \le x)\land \neg T(n,y_1,y_2))$$
and this negation is true (take $y_1=x_0$).
