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A set $X$ is of PA degree relative to a set $Y$ if every infinite $Y$-computable binary tree has an infinite $X$-computable path. A set $X$ is low if $X'$ is computable from $\emptyset'$.

Easy facts:

By the relativized Low basis theorem and using the fact that a low relative to a low is still low, $\emptyset'$ is of PA degree relative to every low set. Of course, if $\emptyset'$ is of PA degree relative to a set $X$, then $X$ is $\emptyset'$-computable.


Are there non-low set $Y$ such that $0'$ is of PA degree relative to $Y$ ?

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up vote 6 down vote accepted

No, by the Arslanov completeness criterion $0'$ is only DNC relative to low sets. And PA implies DNC.

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Thank you for your quick answer. I'm ashamed not to have found it by myself. – Ludovic Patey Jan 10 '14 at 16:47
What is DNC? Thanks – user40919 Jan 11 '14 at 20:07
Diagonally non-computable – Bjørn Kjos-Hanssen Jan 11 '14 at 20:44

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