TITLE: Stronger exact pairs
QUESTION [6 upvotes]: Question: Suppose $\{a_n : n < \omega \}$ is a $<_T$-ascending sequence in $2^{\omega}$. Can we find $x, y \in 2^{\omega}$ such that for every $z \in 2^{\omega}$, the set of reals computable from each one of $x \oplus z$ and $y \oplus z$ is the Turing ideal generated by $\{a_n : n < \omega\} \cup \{z\}$? So taking $z$ to be computable, this implies that $x, y$ form an exact pair for the Turing ideal generated by $\{a_n : n < \omega\}$. 
My thoughts: The usual construction of exact pair has only countably many splitting requirements ($\Phi^{x} = \Psi^{y} = w \implies (\exists n)(w \leq_T a_n)$) but this construction would require continuum many. I suspect such pairs don't exist but I don't have a proof.

REPLY [5 votes]: You could look at Theorem 3.1 of this paper:
On the Σ2-Theory of the Upper Semilattice of Turing Degrees
Author(s): Carl G. Jockusch, Jr. and Theodore A. Slaman
Source: The Journal of Symbolic Logic, Vol. 58, No. 1 (Mar., 1993), pp. 193-204 Published by: Association for Symbolic Logic
3.1 characterizes the possible upper-semi-lattice end-extensions of a countable ideal that can be generated by the ideal and a single new element.  Basically, anything that is consistent as an upper-semi-lattice can be realized.  In particular, for any $x$ and $y$ as above, there is a $z$ such that $z\oplus x$ and $z\oplus y$ are Turing equivalent, but for all $a_n$, $z\oplus a_n$ computes neither $x$ nor $y$.
Your instincts were correct: there is no pair of the described type.