# How much does proving that a special case of a problem is NP-complete tell me about if the general problem is NP-complete?

Define a graph problem as follows. Given a graph $$G$$ and two integers $$c$$ and $$k$$, delete $$k$$ nodes and all edges incident to them, such that, in the remaining graph, every connected component has at most $$c$$ nodes.

What can I say about the time complexity class of the above problem? I can easily prove that under the condition that $$c=1$$, I can reduce Vertex Cover to the problem in polynomial time, and thereby prove that the problem is NP-hard under these parameters.

My work book says that this, accompanied with a polynomial verification, is enough to prove that the problem is NP-complete, but I have a hard time accepting this since we have only observed the problem under a certain set of parameters?

If a problem $$P'$$ is a special case of problem $$P$$, this means that $$P'$$ can be reduced to $$P$$. Therefore, if $$P'$$ is NP-hard, it follows that $$P$$ is NP-hard (because if any problem $$H$$ in NP can be reduced to $$P'$$, then it can also be reduced to $$P$$).
If you also have a proof that $$P$$ is in NP (your proof of polynomial verification; for $$P$$, not just for $$P'$$), then you have a proof that $$P$$ is NP-complete.