I am trying to learn the RiceShapiro theorem on my own. Here is the theorem:Let $$ be a set of computable functions such that the set $R_$ is recursively enumerable. We have $f  $ if and only if there exists a finite function $  $ such that $  f$.I will appreciate if anybody helps me to make the theorem crystal clear.I just want to know what it really means and how I can use it to say that a set is not r.e., for example the set Empty or TOT. 
RiceShapiro theorem in computation theory
computability
Rice-Shapiro greatly helps in proving that a set is not RE, for some sets of the form $R_\Gamma$.E.g. take $A = \{ n \ |\ {\sf dom}(\phi_n) \mbox{ finite} \}$. We have $A=R_\Gamma$ with $\Gamma = \{ \phi_n | {\sf dom}(\phi_n) \mbox{ finite} \}$. If $A$ were RE, since e.g. the always undefined function belongs to $\Gamma$, then by Rice-Shapiro any recursive extension of that would be long to $\Gamma$ as well -- but this is false, since e.g. the identity function  $\notin\Gamma$. Hence $A$ is not RE.The above one is a typical use, where you reach a contradiction by extending a function inside $\Gamma$ to one recursive function outside.We can also use it in the other direction: we reach a contradiction if we can take any function inside $\Gamma$ whose finite restrictions are all outside $\Gamma$. For instance, take $B = \{n \ |\ \forall k\in\mathbb{N}.\ \phi_n(2k)=5 \}$. Clearly $B=R_\Gamma$ with $\Gamma = \{\phi_n \ |\ \forall k\in\mathbb{N}.\ \phi_n(2k)=5 \}$. The function $f(x)=5$ is inside $\Gamma$, but any finite restriction of $f$ can not evaluate to $5$ on all even numbers, hence it must be found outside $\Gamma$. If $B$ were RE, by Rice-Shapiro one such restriction should instead be found inside $\Gamma$. We conclude that $B$ can not be RE.I find Rice-Shapiro easier to use if I think about it as two separate results: One is a monotonicity result, telling us that if $f \in \Gamma$ and $f \subseteq g$ with $g$ partial recursive, then $g \in \Gamma$. I.e. recursive extensions of functions in $\Gamma$ must belong to $\Gamma$ -- otherwise, $R_\Gamma$ is not RE.The other is a compactness result. For any function in $\Gamma$, there must be a finite restriction of it inside $\Gamma$ -- otherwise, $R_\Gamma$ is not RE.Many, many common examples of non-RE sets violate one of these two conditions, making it simple to prove they are not RE. For instance, consider $C = \{n\ |\ \phi_n(3)\neq 4 \land \phi_n(5)=2 \}$, where the notation $\phi_n(3)\neq 4$ is meant to be true when $\phi_n(3)$ is undefined. This immediately violates monotonicity, since I can always extend a (recursive) function which is undefined on $3$ so that it evaluates to $4$ on that point (and still get a recursive function). So $C$ is not RE.