So a common method used to construct non-zero $\omega$-REA *arithmetic* degrees with various properties is to build an $\omega$-REA operator $J$ satisfying the constraints that (for all $X$)


$$\tag{1} J(X') \equiv\_T J(X) \oplus X'$$
$$\tag{2} J(X) >\_T X$$


Inductively, 1 implies that $J(X^n) \equiv\_T J(\emptyset) \oplus X^n$. Thus, together, these constraints ensure that $J(\emptyset)$ isn't arithmetic (if $J(\emptyset) \leq\_T 0^n$ then $J(0^n) \equiv\_T 0^n$).


My question is whether this is fully general, i.e., if $A$ is $\omega$-REA but non-arithmetic is there some $\omega$-REA operator $J$ satisfying 1 and 2 such that $A$ and $J(\emptyset)$ have the same *arithmetic* degree?


Basically, I'm hoping someone will let me know if I'm missing some obvious elementary result or known result before I spend any time trying a hard construction.



Generality of construction for $\omega$-REA arithmetic degrees
So after some careful thought I'm pretty sure it is fully general. Let $A$ be a non-arithmetic $\omega$-REA set and $K(X)$ some $\omega$-REA operator satisfying 1 and 2 (see Odifreddi volume 2 XIII.3.1 for an existence proof). We now define $J(X)$ so that $J(X) \equiv\_T X \oplus A$ if $X = \emptyset^n$ for some $n$ and otherwise make $J(X) \equiv\_T K(X) \oplus A$.


**Claim 1**: If $J(X)$ behaves as indicated then $J(X)$ satisfies (1), (2) and $J(\emptyset) = A$.


*Pf:* Since $A$ is non-arithmetic we trivially have (1) satisfied for any arithmetic $X$. If $X$ not arithmetic then $J(X) \geq\_T K(X) >\_T X$. As the jump is injective to verify (2) it is enough to note that it clearly holds separately for both $K(X) \oplus A$ and $X \oplus A$.$\square$


**Claim 2**: The desired operator $J(X)$ exists if there is a $\Pi^0\_2$ class $P$ with $P = \lbrace X \mid (\exists k)(X = \emptyset^k) \rbrace$


*Pf:* Given such a $\Pi^0\_2$ class $P(X) \iff \forall n \exists m Q(X\restriction\_m, n, m)$ we build an r.e. operator $R(X)$ by enumerating $n$ when we've found $m$ for each $n' \leq n$ ensuring that $$R(X) =^{\*} \begin{cases}
\emptyset & \text{if } (\exists k)(X = \emptyset^k) \\
\omega & \text{ otherwise} \\
 \end{cases}$$


We let $J^{[1]}(X) = R(X)$. Obviously, the column-wise sum of $K(X)$ and $A$ is $\omega$-REA in $X$ so we simply modify this by only enumerating elements into the $K(X)$ part if $(\exists z)(z \notin R(X))$.
$\square$


We now define the computable predicate $Q(\tau, m, n)$.


Fix a totally total (i.e. total for all $X$) computable functional $\rho$ such that $\rho(Y') = Y$ and let $\tau^{k} = \rho^{k}(\tau)$ where $\rho^{k}$ denotes the application of $\rho$ $k$ times with $\rho^0$ the identity (WLOG $n \leq \lvert \tau^{n+1}\rvert \leq \lvert \tau^{n}\rvert \leq \lvert \tau\rvert = m$).


Let $$k(n,m) = \max\_{k < m}\,\,\,\lvert \tau^{k}\rvert \geq n \land (\forall j < k) )(\forall x < n) (\phi\_x(\tau^{j}; x)\downarrow\_m \iff \tau^{j-1}(x) = 1 $$


Thus $k(n,m)$ is our guess at the maximum number $k$ such that $X = Y^{k}$ for some $Y$ and $\tau^{k(n,m)}$ our approximation to $Y$. Using this we now define


$$Q(\tau, m, n) \iff (\exists k \leq k(n,m))(\forall x < n)(\tau^k(x) = 0)$$


**Claim 3**: If $P(X)$ is defined to be $\forall n \exists m Q(X\restriction\_m, n, m)$ then $P(X) \iff (\exists k)(X = \emptyset^k)$.


*Pf:* If $X = \emptyset^k$ the claim is straightforward. $\tau^{k}$ will be identically $0$ and any $m > k$ large enough to witness the convergence all the finitely many computations examined in checking $Q(X\restriction\_m, n, m)$ will suffice (such .


For the other direction suppose that $k$ is the maximum such that for some $Y$, $Y^{k} = X$ (the existence of such a $k$ follows from [this question](https://mathoverflow.net/questions/378631/infinite-descending-chain-of-turing-jumps-with-equality)). Thus, for all sufficiently large $n$, $k(n,m) \leq k$ as $m > n$ sees the failure of $\rho^{k+1}(X)' = \rho^{k}(X)$. By hypothesis no $\rho^{q}(X) \neq \emptyset$ for any $q \leq k$ so if $n$ is larger than all the witnessing locations then $\exists m Q(X\restriction\_m, n, m)$ fails to hold and thus $P(X)$ fails to hold. $\square$




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[1]: Specifically if we denote the $n$-th column of $J(X)$ by $J^{[n]}(X)$, we'll build 
$$J^{[0]}(X) = X,J^{[1]}(X) =^{\*} \emptyset, J^{[n+2]}(X) = K^{[n+1]}(X) \oplus A^{[n+1]} \text{ if } (\forall k)(X \neq \emptyset^k) \text{ and } \tag{a}$$
$$J^{[0]}(X) = X,J^{[1]}(X) =^{\*} \omega, J^{[n+2]}(X) = \emptyset \oplus A^{[n+1]} \text{ if } (\exists k)(X = \emptyset^k) \tag{b}$$


