Given a finite subset $S$ not containing the identity element in a residually finite group $G$, does there always exist a normal subgroup of $G$ which has finite index (in $G$) and which avoids $S$? (If $S$ is a singleton, this is of course the definition of a residually finite group.)
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Yes, take the intersection of the normal subgroups $N_1, N_2, ..., N_k$ of finite index avoiding elements $x_1,x_2,...,x_k$ of your set. It is normal and of finite index (at most the product of indices of $N_i$). 


The answer is yes. Since $G$ is residually finite, given $g\in S$ there is a finite group $F_g$ and a homomorphism $\phi_g: G\to F_g$ whose kernel does not contain $g$. Now $S$ is disjoint from the kernel of the product of these homomorphisms $G\to \prod_{g\in S}F_g$, and the target is a finite group. More generally if $C$ is a class of groups closed under finite products, and $G$ is residually $C$, then $G$ is fully residually $C$. 

