Is a subset of a NP-complete language also NP-complete?

Question

For example, we know that $SAT$ is NP-complete. However, what if we have a set $subSAT \subset SAT$. Is $subSAT$ NP-complete?

What if we have a set $numSAT$ where $numSAT = \{ x \in SAT \; | \; |x| < 20\}$? Is $numSAT$ NP-complete?

Answer 1

The answer to the question in the title is "No". For example, the empty language is a subset of any language, and is decidable in constant time, of course.

So it really depends on what subset of SAT you want to take. With your definition, $numSAT$ is in $P$, as it is a finite language (any finite language is decidable in constant time). An example of a proper subset $L$ of $SAT$ which is NP-complete is the set of boolean formulas in conjunctive normal form, where each clause has size at most $3$: this is the well-known $3$SAT problem.

Note that the converse question "If $L$ is NP-complete, is any $L'\supset L$ NP-hard?" also has a negative answer, since the full language $\Sigma^*$ is decidable in constant time. This simply says that the complexity measure does not behave well with set inclusion.