Undecidable completion of undecidable theory, and pairs of RCF Given an undecidable collection of first-order sentences, is there necessarily a complete undecidable theory containing it? A direct attempt to prove it seems to require some control over the completions which need not exist, but I don't see a counterexample.
On a more concrete side, Macyntire proved that the theory of all pairs of real closed fields is undecidable. I am interested to know if there is a particular pair of real closed fields with undecidable theory.
 A: Even given Stefan's reformulation, the answer is still "no."
Let $L$ be a language consisting only of countably many constant symbols $c_0$, $c_1$, $c_2$, . . . and a single unary relation symbol $U$. Let $\phi$ be the sentence asserting that $U$ holds of exactly one element in the universe: $\phi\equiv\exists ! x(U(x))$.
Now we just adapt Gabriel's example. Let $A$ be any undecidable set. Let $S=\lbrace\neg U(c_i): i\in A\rbrace$, and let $T$ be the closure of $S\cup\lbrace\phi\rbrace$ under implication. Then $T$ is clearly undecidable. However, any complete extension of $T$ is decidable.
A: My recollection is that Macintyre proved there are $2^{\aleph_0}$ complete theories of pairs
$(K,L)$ where $L\subset K$ are real closed fields. This is in his thesis but I don't think
he published it anywhere else.  There are later papers of Francoise Delon and Walter Bauer
that develop this further.  On the other hand there is an earlier theorem of Robinson's that
if $L\subset K$ are real closed and $L$ is dense in $K$ then the theory of $(K,L)$ is decidable.   
A: A counterexample for the first question:
Assume a language with no predicates, functions or constants.  Take an undecidable set $A \subseteq \mathbb{N}$.  The set Γ of all sentences whose length is in A is certainly undecidable.  However all theories in that language are decidable.
