Approximating Turing reducibility with finite subsets

Question

Let $A$ and $B$ denote infinite subsets of $\omega$. Does there exist a quasi-ordering $\leq$ on $[\omega]^{<\omega}$ such that either:

1. $A \leq_\mathrm{T} B$ iff for all finite $x \subseteq A$, there exists some $y \subseteq B$ such that $x \leq y$, or;
2. $A \leq_\mathrm{T} B$ iff for almost all finite $x \subseteq A$, there exists some $y \subseteq B$ such that $x \leq y$?

Here, "almost all" means for all but finitely many finite subsets of $A$.

Answer 1

No such relation can exist (whether it's a quasi-ordering or not).

Since $\;\omega\leq_\mathrm{T}\omega,\;$ we would have that for all (or almost all) finite $x \subseteq \omega$, there exists some $y \subseteq \omega$ such that $x \leq y.$

But then for any set $A,$ for all (or almost all) finite $x \subseteq A$, there exists some $y \subseteq \omega$ such that $x \leq y.$ It would follow that $A\le_{\mathrm T}\omega,$ so $A$ is recursive. So every set of natural numbers would be recursive, which is false.

Answer 2

As Mitchell Spector has observed, the answer to your question is negative. However, there is an interesting phenomenon of a vaguely-similar nature that occurs if we shift from Turing to enumeration reducibility, and consider a single reduction operator instead of the whole reducibility notion:

We say $X\le_eY$ iff there is a computable set $\Phi$ - called an enumeration operator - of pairs $\langle A, B\rangle$ of finite sets of natural numbers such that $$\forall A\in[\omega]^{<\omega}[A\subseteq X\iff \exists B\subseteq Y(\langle A, B\rangle\in\Phi)].$$ Now suppose we fix an enumeration operator $\Phi$. This gives rise to a binary relation (usually not transitive though) $\triangleleft_\Phi$ on $[\omega]^\omega$ such that $X\le_eY$ via $\Phi$ iff for each finite $A\subseteq X$ there is some finite $B\subseteq Y$ with $A\triangleleft_\Phi B$.

Note, though, that it's crucial to this picture that enumeration reductions use only positive information about the set $Y$, and conversely only make positive conclusions about the set $X$. Turing reductions use both positive and negative information to get both positive and negative information, and so it's better to think about finit partial maps $\omega\rightarrow 2$ than finite sets of natural numbers if we want to whip up an analogous picture for (a single) Turing reduction.