Is every limit computable set in the Ershov hierarchy?

Question

A set is limit computable, or $\Delta^0_2$, if its characteristic function is equal to $\lim_{s\rightarrow\infty}g(x,s)$ for some total computable function $g$. And given a computable ordinal $\alpha$ and a path $P$ through Kleene's $O$, a set is $\alpha$-c.e. if its characteristic function is equal to $\lim_{s\rightarrow\infty}g(x,s)$ for some total computable function $g$ such that there exists a total computable function $o:\mathbb{N}^2\rightarrow P$ such that $o(x,0)$ is a notation for $\alpha$, $o(x,s+1)\leq_O o(x,s)$ for all $s$, and $o(x,s+1)<_Oo(s,x)$ whenever $g(s+1,x)\neq g(s,x)$. The latter is known as the Ershov hierarchy.

Now I'm wondering if the Ershov hierarchy exhausts limit computable sets. In one sense the answer is a trivial yes: for every limit computable set $X$ there exists a path $P$ through Kleene's $O$ such that $X$ is $\omega^2$-c.e. And also there exists a path $P$ through Kleene’s O such that every limit computable set is $\omega^3$-c.e. But my question is, for any given maximal path $P$ through Kleene's $O$, is every limit computable set an $\alpha$-c.e. set for some computable ordinal $\alpha$?

Answer 1

No. Not only is it not the case that every maximal path in $O$ yields an Ershov hierarchy which exhausts the $\Delta^0_2$ sets, Yuri Ershov showed that there’s a great many maximal paths in $O$ which fail to have this property. See theorem 4.6 of this survey article.

To wit, $P$ be a $\Pi^1_1$ maximal path in $O$, and suppose that the Ershov hierarchy with respect to $P$ exhausted all $\Delta^0_2$ set. Fix an enumeration of $\Delta^0_2$ sets, and let $h:\mathbb{N}\rightarrow P$ be defined by, $h(n)$ is the $O$-least element $a$ of $P$ such that the $a^{th}$ $\Delta^0_2$ is $a$-c.e. Then $h$ is total and thus hyperarithmetical, which by $\Sigma^1_1$-boundedness implies that the range of $h$ is bounded, yielding a contradiction. Therefore, the Ershov hierarchy with respect to any $\Pi^1_1$ maximal path in $O$ does not exhaust the $\Delta^0_2$ sets.