Why does $L\subseteq \textbf{P} \cap \textbf{NP}$ is $\textbf{NP}$-complete imply $\textbf{NP} = \textbf{P}$?

Question

If I show that a language $L$ is contained in $\textbf{P}$ and $\textbf{NP}$ and I know that the language is $\textbf{NP}$-complete, why did I proof that $\textbf{P} = \textbf{NP}$?

Answer 1

Because you have then showed that $L$ is an $\textbf{NP}$-complete language which, since $L \in \textbf{P}$, is decidable in poly-time. Since any other language $L' \in \textbf{NP}$ is efficiently reducible to $L$ (because of $\textbf{NP}$-completeness), $L' \in \textbf{P}$ as well. It follows that $\textbf{NP} \subseteq \textbf{P}$ (and the other inclusion is trivial).