I seem to recall once hearing a result to the effect that $\emptyset^{(\omega)}$ was the double jump of some other degree, but could not be the triple jump of any degree. However I'm unable to find the exact result. Does anyone know what I might be thinking of (or what is actually known about jump inversion on $\emptyset^{(\omega)}$, if I'm remembering this completely wrong)?
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3 Answers
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This doesn't exactly answer your question but...
If $A$ is any upper bound for the arithmetic degrees then $0^{(\omega)}$ is recursive in $A^{\prime\prime}$. Enderton and Putnam proved that there upper bounds with $A^{\prime\prime}=0^{(\omega)} $
Dave
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3$\begingroup$ But this does answer my question: the question was what result I was thinking of, and I'm fairly sure this was the result. $\endgroup$ Commented May 16, 2011 at 2:32
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As far as actual jump inversion goes, any degree $X \geq 0^{(n)}$ is the $n$th jump of some other degree. An easy way to see this is to apply Friedberg jump inversion relative to $0^{(n-1)}$, then relative to $0^{(n-2)}$, and so on down to $0$. The theorem is also true through transfinite iterates of the jump: if $X \geq 0^{(\alpha)}$, then $X$ is the $\alpha$th jump of some degree. The general version of this theorem for any $\alpha$-REA operator is due to Jockusch and Shore (1984).
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I suspect you were looking for the stuff answered above.
But, if you want arithmetic jump inversion you can have that too (Simpson ..nice presentation in Odifreddi II). Essentially what you would expect from r.e. jump inversion.
If $C$ is $\omega$-REA in $0^\omega$ then there is $\omega$-REA $A$ with $A'$ of same arithmetic degree as $0^\omega$ join $C$. This fails for T degrees but you can show that for any $\omega$-REA operator J there is an $\omega$-REA A with J(A) Turing equivalent to $0^\omega$.
