In a book on Type theory I am working through, there is a definition for subterms where the third case covers lambda abstractions and is given by $$Sub((\lambda x. M)) = Sub(M) \cup \{(\lambda x. M)\}$$ but I would have thought it should be $$Sub((\lambda x. M)) = Sub(M) \cup \{(\lambda x. M)\} \cup \{x\}.$$
Why isn't the set containing $x$ included, since $Sub(x)$ is given as $\{x\}$ in case 1 (basis)?
For example, in the identity function, $$\lambda x. x,$$ the second $x$ is considered a subterm, but the first one (after $\lambda)$ is not. Just wondering why not.