TITLE: Consistent r.e. extensions of non r.e. theories.
QUESTION [5 upvotes]: Let $\mathcal{L}$ be some first-order language, and $T$ be a consistent set of formulas of $\mathcal{L}$ which is not recursively enumerable.
Under what conditions will there be $T'\supset T$ such that $T'$ is consistent and recursively enumerable?

REPLY [2 votes]: Let $A = T \cup R$ where $R$ is any set of refutable statements from $T$ (i.e., $T \vdash \lnot \varphi$ for all $\varphi$ in $R$).  A simple necessary condition for such a $T^{\prime}$ to exist is that $A$ is not recursively (computably) enumerable.  If it were, then given a r.e. (c.e.) $T^{\prime} \supseteq T$, we could recursively enumerate all statements of $T$ by listing only the ones in $T^{\prime}$ that also appear in $A$ (i.e., $T = T^{\prime} \cap A$).  The reason for this equality is that $T^{\prime}$ is an extension of $T$ so it must contain all the statements in $T$ but then it cannot include any in $R$ by its consistency.
Now let $I$ be the set of all statements that are independent of $T$ (i.e. all $\varphi$ such that $T \nvdash \varphi$ and $T \nvdash \lnot \varphi$).  In the case that $T$ is  deductively closed meaning $T \vdash \varphi$ implies that $\varphi \in T$, a corollary of the above result is that $I$ cannot be the complement of a recursively enumerable set.  This is the instance of letting $R$ be the set of all statements refutable from $T$.