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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.

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    $\begingroup$ I suspect that for $g$ sufficiently Cohen generic, the real $z=(x\Delta g)\oplus (y\Delta g)$ will fail to compute $x$ (or $y$) even from finitely many of the $a_i$s; since $z\oplus x$ and $z\oplus y$ each compute $x\oplus y$, this would give the expected negative answer. But I don't yet see how to show this (in particular, the naive forcing analysis just gets us $(x\Delta y)\oplus a_1\oplus ...\oplus a_n\equiv_Tx\oplus y$ for some $n$, which is entirely possible). $\endgroup$
    – Noah Schweber
    Commented May 20, 2019 at 9:15

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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.

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