TITLE: Example of sets $A, B$ such that $A', B'$ are Turing equivalent but $A, B$ are not.
QUESTION [5 upvotes]: I have been wondering if the following statement is true,
$$
A\equiv_TB\iff A'\equiv_TB'
$$
where $A, B\subseteq\omega$ and $A'$ denotes the Turing jump of $A$. I have been able to show the forward direction, but have been unable to show the converse. I am starting to think that the converse fails and have been looking for an example. Any suggestions?

REPLY [2 votes]: Similar to @Alex's answer, there are also low sets. A set $A\subseteq \omega$ is low if $A' \leq 0'$ (so that $A' \equiv_{T} 0'$). Notice that any low $0<A<0'$ would then be a counterexample to the converse of your statement. The Kleene-Post finite-extension construction of incomparable degrees builds two such sets.