This was asked at MSE without success. Granted, a bounty is still ongoing there, but it doesn't look like it will be answered.

For $A\subseteq\omega$ let $\tilde{A}=\{e: \varphi_e^A\mbox{ is total and }\exists c(\varphi_e^A\simeq\varphi_c)\}$. A priori $\tilde{A}$ is $\Sigma^0_3(A)$ ("$\varphi_e^A$ is total and there is some $c$ such that on all inputs we eventually see agreement between $\varphi_c$ and $\varphi_e^A$"). However, this bound isn't sharp in general: if $A$ is sufficiently Cohen generic then $\tilde{A}$ is $\Pi^0_2(A)$ (of course we can't do better than this: $\tilde{A}$ is always $\Pi^0_2(A)$-hard).

However, a fair amount of genericity (at a glance, $2$-genericity) is needed for that argument. This raises the question of how hard it must be to compute a real satisfying "$\tilde{A}$ is not $\Sigma^0_3(A)$-complete." Specifically:

Does every $\Delta^0_2$ set $A$ satisfy "$\tilde{A}$ is $\Sigma^0_3$-complete"?

Genericity-based arguments almost certainly won't be useful here, since there aren't even any $\Delta^0_2$ weak $2$-generics.

This was asked at MSE without success. Granted, a bounty is still ongoing there, but it doesn't look like it will be answered.

For (noncomputable) $A\subseteq\omega$ let $\tilde{A}=\{e: \varphi_e^A\mbox{ is total and }\exists c(\varphi_e^A\simeq\varphi_c)\}$. A priori $\tilde{A}$ is $\Sigma^0_3(A)$ ("$\varphi_e^A$ is total and there is some $c$ such that on all inputs we eventually see agreement between $\varphi_c$ and $\varphi_e^A$"). However, this bound isn't sharp in general: if $A$ is sufficiently Cohen generic then $\tilde{A}$ is $\Pi^0_2(A)$ (of course we can't do better than this: $\tilde{A}$ is always $\Pi^0_2(A)$-hard).

However, a fair amount of genericity (at a glance, $2$-genericity) is needed for that argument. This raises the question of how hard it must be to compute a real (nontrivially) satisfying "$\tilde{A}$ is not $\Sigma^0_3(A)$-complete." Specifically:

Does every noncomputable $\Delta^0_2$ set $A$ satisfy "$\tilde{A}$ is $\Sigma^0_3$-complete"?

Genericity-based arguments almost certainly won't be useful here, since there aren't even any $\Delta^0_2$ weak $2$-generics.

This was asked at MSE without success. Granted, a bounty is still ongoing there, but it doesn't look like it will be answered.

For $A\subseteq\omega$ let $\tilde{A}=\{e: \varphi_e^A\mbox{ is total and }\exists c(\varphi_e^A\simeq\varphi_c)\}$. A priori $\tilde{A}$ is $\Sigma^0_3(A)$ ("$\varphi_e^A$ is total and there is some $c$ such that on all inputs we eventually see agreement between $\varphi_c$ and $\varphi_e^A$"). Moreover, $\tilde{\emptyset}$ is $\Sigma^0_3$-complete. However, this bound isn't sharp in general: if $A$ is sufficiently Cohen generic then $\tilde{A}$ is $\Pi^0_2(A)$ (of course we can't do better than this: $\tilde{A}$ is always $\Pi^0_2(A)$-hard).

However, a fair amount of genericity (at a glance, $2$-genericity) is needed for that argument. This raises the question of how hard it must be to compute a real satisfying "$\tilde{A}$ is not $\Sigma^0_3(A)$-complete." Specifically:

Does every $\Delta^0_2$ set $A$ satisfy "$\tilde{A}$ is $\Sigma^0_3$-complete"?

Genericity-based arguments almost certainly won't be useful here, since there aren't even any $\Delta^0_2$ weak $2$-generics.

This was asked at MSE without success. Granted, a bounty is still ongoing there, but it doesn't look like it will be answered.

For $A\subseteq\omega$ let $\tilde{A}=\{e: \varphi_e^A\mbox{ is total and }\exists c(\varphi_e^A\simeq\varphi_c)\}$. A priori $\tilde{A}$ is $\Sigma^0_3(A)$ ("$\varphi_e^A$ is total and there is some $c$ such that on all inputs we eventually see agreement between $\varphi_c$ and $\varphi_e^A$"). However, this bound isn't sharp in general: if $A$ is sufficiently Cohen generic then $\tilde{A}$ is $\Pi^0_2(A)$ (of course we can't do better than this: $\tilde{A}$ is always $\Pi^0_2(A)$-hard).

However, a fair amount of genericity (at a glance, $2$-genericity) is needed for that argument. This raises the question of how hard it must be to compute a real satisfying "$\tilde{A}$ is not $\Sigma^0_3(A)$-complete." Specifically:

Does every $\Delta^0_2$ set $A$ satisfy "$\tilde{A}$ is $\Sigma^0_3$-complete"?

Genericity-based arguments almost certainly won't be useful here, since there aren't even any $\Delta^0_2$ weak $2$-generics.