# Set of $\mathsf{NP}$-hard languages closed under set inclusion?

As the title says, my question is whether the set of $\mathsf{NP}$-hard languages is closed under set inclusion, i.e. whether for any $\mathsf{NP}$-hard language $L$, all subsets of $L$ are also $\mathsf{NP}$-hard. This question is related since $\emptyset$ is not $\mathsf{NP}$-hard as there is nothing we could map "yes"-instances to and $\emptyset \subseteq L$ for all $\mathsf{NP}$-hard $L$.

However, what about non-trivial subsets, i.e. languages $L'$ of the form $\emptyset \neq L' \subsetneq L$ for $\mathsf{NP}$-hard $L$? We know that $\emptyset \neq 2SAT \subsetneq SAT$ and while $SAT$ is $\mathsf{NP}$-hard, $2SAT$ is in $\mathsf{P}$. This suggests to me that whether the set of $\mathsf{NP}$-hard is closed under "non-trivial" inclusions depends on whether $\mathsf{P} = \mathsf{NP}$. Am I mistaken here?

In one sentence, my question is this: is the set of $\mathsf{NP}$-hard languages closed under nontrivial set inclusion (1) assuming $\mathsf{P} = \mathsf{NP}$ and (2) assuming $\mathsf{P} \neq \mathsf{NP}$?

• That's what I figured, and I'm really surprised by it - it seemed "obvious" to me that regular languages should not be NP-hard. Do you know a proof that regular languages are not NP-hard assuming P $\neq$ NP? – G. Bach May 19 '14 at 15:16