[Solved]: NP complete language having no Polytime decidable superset

No, assuming P$,neq,$NP. Let $L$ be any NP-complete language over alphabet $Sigma$ and let $N = {ninmathbb{N}mid Lcap Sigma^nneq Sigma^n}$, i.e., $N$ is the set of lengths such that at least one word of length $n$ is not in $L$. Note that $N$ must be infinite. If not, it has some maximum element $m$ and $L = big(Lcap Sigma^{leq m}big) cup Sigma^{>m}$, which is a regular language, since $LcapSigma^{leq m}$ is finite. So, in particular, $L$ is in P, contradicting the assumption that it is NP-complete (under the assumption that P$,neq,$NP). But now pick any $nin N$ and let $L_n = big(Lcap Sigma^{leq n}big)cupSigma^{>n}$. $L_n$ is a strict superset of $L$ (because, for any $n<n’in N$, $L_n$ contains all strings of length $n’$ but $L$ doesn’t) and $L_nin,$P by the same argument as above.