Are the Turing degrees of truth sets cofinal? Godel’s Completeness Theorem, in Henkin’s version at least, says that every consistent countable set of first-order sentences has a model.  Based on that, let us call a set of natural numbers a truth set if it is the set of Gödel numbers of some consistent countable set of first-order sentences.
My question is, are the Turing degrees of truth sets cofinal in the set of all Turing degrees?  If not, what is the supremum of the set of all Turing degrees of truth sets?  I imagine it would be pretty large if it is exists at all.
 A: In fact, for any $A\subset\mathbb{N}$, we can find a theory $T$ such that $A$ is one-one reducible to $T$.
Consider the language which have an unary predicate symbol $P$ and a recursive enumeration of countable constant symbols $\langle c_n |n<\omega\rangle$. For each $A\subset\omega$, define $T$ be a theory given by the following set of sentences:


*

*$c_i\neq c_j$ for all $i\neq j$,

*$P(c_n)$ if $n\in A$ and

*$\lnot P(c_n)$ if $n\notin A$.


$(\omega, A, n)_{n\in\omega}$ is a model of $T_A$, so $T_A$ is consistent. Furthermore, we can see that $n\in A$ iff $T\vdash P(c_n)$. Therefore, a recursive map $ n\mapsto \ulcorner P(c_n)\urcorner$ witnesses $A \le_1 T$.

In addition, we can do a bit more: we can construct a consistent theory $T$ such that $A\le_1 T$ and $T\le_T A$. Consider the extension $T_A'$ of $T_A$ obtained by adding two axiom schemes which state "there are infinitely many $x$ such that $P(x)$" and "there are infinitely many $x$ such that $\lnot P(x)$". We can see that $T_A'$ allows quantifier elimination. Since $T_A'$ decides every atomic formula, $T_A'$ is complete. Moreover, we can find a model of $T_A'$ easily, so $T_A'$ is consistent.
By the same argument, $A \le_1 T_A'$ holds. Conversely, we can see that $T_A'$ is $A$-recursive as it is a complete theory given by a set of $A$-recursive axioms. Hence $T_A'\le_T A$. As a corollary, every Turing degree has a consistent theory which is Turing-equivalent to this degree.
