I'd like to show that the language DISCONNECTED-SET, defined by $$ \{\langle G, k\rangle | \text{$G$ is an undirected graph that contains a disconnected set of size $k$}\},$$ is NP-complete. (A disconnected set $S$ is defined as a set of vertices such that for every pair of vertices $u,v \in G$, if there is no edge between $u$ and $v$, then at least one of the vertices is in $S$.)I'm under the assumption that it could be done by reduction from CLIQUE. I've made the following observations (which I'm unable to utilize, though):If we have a graph $G$, then the maximum disconnected set of its complement $G'$ has a size equal to the number of nodes in $G$ that belong to connected components of size at least 2 in $G$.If we have isolated nodes in $G$, the sizes of the disconnected sets in $G'$ are unchanged.May I just ask for a tiny tip that would get me going in the right direction, not a complete proof. Thank you. 
NP-completeness of DISCONNECTED-SET
complexity theory;np complete;np hard
This is just VERTEX-COVER in disguise. The reduction from CLIQUE is also simple  a set is disconnected iff its complement is a clique.