For a given set of numbers $A$, let $O^A$ be the hyperjump of $A$. It is possible to iterate inductively the hyperjump of a set, through the computable ordinals, in a way that the $\alpha$-th hyperjump is computably stronger than any $\beta$-th hyperjump for $\beta < \alpha$. It is even possible to keep iterating it through the ordinals which becomes computable in a previous iterations of the hyperjump. Formally, let us do the following inductive definition of an order $<$ on the natural numbers, together with sets $J\_a^A$ for any $a$ in the field of the order $<$ (these elements $a$ are meant to be notations for ordinals) :


* Let $J\_1^A$ be the set $A$ and $J\_2^A$ be the set $O^A$. Define $1 < 2$.
* If $a$ is in the field of $<$, let $J\_{2^a}^A = O^{J\_a^A}$. Define $a < 2^a$ and $b < 2^a$ for any $b < a$.
* If $a$ is in the field of $<$ and if $e$ is the code of a computable functional such that $\varphi\_e(0, J^A\_a) = a$ and $\varphi\_e(n, J^A\_a) < \varphi\_{e}(n+1, J^A\_a)$, then $J^A\_{3^e5^a}$ is the disjoint union of the sets $J^A\_{m\_i}$ for $m\_i = \varphi\_e(i, J^A\_a)$. We define $b < 3^e5^a$ for any $b < \varphi\_e(i, J^A\_a)$ for some $i$.


There is a smallest ordinal $\lambda^A$ which does not have a corresponding code in the field of $<$. It seems clear (even though I haven't gone through the details carefully) that $\lambda^A$ should be recursively inaccessible.


Sacks showed that for any countable admissible ordinal $\alpha$, there is a set $A$ such that $\alpha$ is the smallest ordinal non-computable ordinal in $A$.


My question is : Is it known/true that any countable recursively inaccessible ordinal is of the form $\lambda^A$ for some set $A$ ? (with $\lambda^A$ defined as above).



On a characterization of the recursively inaccessible ordinals
Well assuming that $\lambda^A$ is always the first recursively admissible which is bigger than $\omega\_1^A$, which I think should be true, I think my question is after all not so interesting:


Either there is a largest recursively inaccessible smaller than $\omega\_1^A$, in which case $\lambda^A$ is a successor in the recursively inaccessible ordinals, or $\omega\_1^A$ is a limit of recursively inaccessible. In this case it is itself also recursively inaccessible (as an admissible being a limit of admissible). Thus $\lambda^A$ is also a successor in the recursively inaccessible ordinals.


It would follow that $\lambda^A$ is never meta-recursively inaccessible : a recursively inaccessible being a limit of recursively inaccessible.


