The full problem is as follows :
If $T \supseteq {\Gamma}_{N}$ is a recursive theory, then the deductive closure of $T$ is not recursive. Use this and the fact that, if $T$ is a recursive theory such that the deductive closure is complete then $\{ [\phi] : T \vdash \phi \}$ is recursive, to show that if $T \supseteq {\Gamma}_{N}$ is a recursive theory, then the deductive closure of $T$ is incomplete.
Just note that ${\Gamma}_{N}$ is the first-order axiomatization of arithmetic as given usually.
I am really straggling to prove this statement. Any recommendations?
Thank you in advance.
The deductive closure $Ded(T)$ of $T$ is definitely recursively enumerable (do you see why?). So suppose it were complete. That means that for every $\varphi$, either $\varphi\in Ded(T)$ or $\neg\varphi\in Ded(T)$.
Since $Ded(T)$ is recursively enumerable, there is a Turing machine $\Phi$ such that $\Phi$ halts on input $[\varphi]$ iff $\varphi\in Ded(T)$ (where $[\cdot]$ denotes the Godel number function). Assuming $Ded(T)$ is complete, do you see how to turn this into a decision procedure for $Ded(T)$?
HINT: Think about what happens if I run $\Phi([\varphi])$ and $\Phi([\neg\varphi])$ in parallel . . .
Note: you need to assume, of course, that $T$ is consistent. The deductive closure of an inconsistent theory is definitely recursive. :P