The existence of a high arithmetic degree (meaning the degrees induced by the notion of relative arithmetic definability) below $0^\omega$ can be established by using Harrington/Simpson's construction of a uniformly low $\omega$-REA operator, i.e., $J(X)$ which satisfies $X <_a J(X) <_a X^{(\omega)} \equiv_T (J(X))^{(\omega)}$ and combining it with Simpson's inversion/completion theorem for $\omega$-REA operator operators to give us an $\omega$-REA set $A$ such that $J(A) \equiv_T 0^{(\omega)}$. Thus, $A^{(\omega)} \equiv_T 0^{(\omega + \omega)}$.
Is there any other construction of a set $X$ with $X <_a 0^{(\omega)}$ and $X^{(\omega)} \equiv_a 0^{(\omega + \omega)}$? If we don't demand $X$ be arithmetic in $0^{(\omega)}$ we can mimic the usual jump-inversion style argument by simply alternating between forcing arithmetic statements and coding elements of $0^{(\omega + \omega)}$. However, if $X <_a 0^{(\omega)}$ that method would produce $X$ satisfying $X \oplus 0^{(\omega)} \equiv_a 0^{(\omega)}$ so we somehow need to ensure that the jump is doing the work.
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Recall that the arithmetic degrees are the equivalence classes induced by the relation $X \leq_a Y$ which holds iff $(\exists n)(X \leq_T Y^{(n)})$ iff $X$ is definable from $Y$ by an arithmetic formula. The arithmetic jump of $X$ is just $X^{(\omega)}$.
