Prove that the set $\{x|\varphi_x \text{is total}\}$ is productive

Question

It's easy to show that the complement $\bar{T}=\{x|\varphi_x \text{is not total}\}$ is productive, since $\bar{K}$ is many-one reducible to it, but what about $T$?

Answer 1

Using s-m-n theorem find a function $f(x)$ with $$ \varphi_{f(x)}(y) = \begin{cases} \uparrow,& \text{if } \varphi_x(x)\!\downarrow \text{in at most $y$ steps}; \\ 0,& \text{otherwise.} \end{cases} $$

If $x \in K$, then $\varphi_{f(x)}$ diverges starting from the number of steps in the computation of $\varphi_x(x)$, and hence it is not total. If $x \notin K$, then $\varphi_{f(x)}$ is constant zero function and hence is total. This shows that $\overline{K} \leqslant_m T$. In fact, $T$ is known to be $\Pi_2^0$-complete, and since $\overline{K}$ is $\Pi_1^0$ it must be reducible to $T$.

Also you can show it directly by the definition of a productive set. Given an r.e. subset $W_x \subseteq T$, we need to effectively produce an index of a total function not in $W_x$. The standard idea here is to use diagonalization:

• set $t = 0$ and start enumerating $W_x$;
• when (if ever) a new element $e$ appears in $W_x$ define $f(t) := \varphi_e(t) + 1$ and set $t := t + 1$;
• while waiting for some $e \in W_x$ let $f(t) = 0$ and set $t := t + 1$;

The construction above is uniform and effective and so for some $p(x)$ we have $\varphi_{p(x)} = f$. Note that $\varphi_e$ is total for $e \in W_x$ and hence $\varphi_e(t)$ always converges, so $f(t)$ is a total function - if $W_x$ is infinite it is clear, but if $W_x$ is finite (or empty) while (endless) waiting for a new element is taking place the last bullet makes $f$ total. So $p(x) \in T$ for all $x$ such that $W_x \subseteq T$. If $p(x) \in W_x$, then at some stage we define $\varphi_{p(x)}(t) = f(t) = \varphi_{p(x)}(t) + 1$, and so $p(x) \in W_x \setminus T$, a contradiction.