Jump Inversion of Arithmetic - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T10:36:38Z http://mathoverflow.net/feeds/question/65099 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/65099/jump-inversion-of-arithmetic Jump Inversion of Arithmetic Henry Towsner 2011-05-16T01:32:10Z 2011-05-18T12:46:21Z <p>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)?</p> http://mathoverflow.net/questions/65099/jump-inversion-of-arithmetic/65102#65102 Answer by Dave Marker for Jump Inversion of Arithmetic Dave Marker 2011-05-16T02:20:36Z 2011-05-18T12:46:21Z <p>This doesn't exactly answer your question but...</p> <p>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 $A^{\prime\prime}=0^{(\omega)}$</p> <p>Dave</p> http://mathoverflow.net/questions/65099/jump-inversion-of-arithmetic/65110#65110 Answer by Andrew Marks for Jump Inversion of Arithmetic Andrew Marks 2011-05-16T04:29:33Z 2011-05-16T04:29:33Z <p>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).</p>