TITLE: Operators with compact resolvent
QUESTION [5 upvotes]: This should be a basic, or even stupid, question, but I am really confused, and I cannot find any webpage that addresses my question.
From wikipedia (https://en.wikipedia.org/wiki/Resolvent_formalism), an operator $A$ has compact resolvent iff $(A - zI)^{-1}$ is compact for some $z$.
My confusion is that, compact operators cannot be invertible if the domain is infinite dimensional, but clearly $(A - zI)^{-1}$ is invertible by definition. Then this definition would not make sense!

REPLY [5 votes]: Indeed, when the resolvent is compact and the space is not finite dimensional, the operator must be unbounded. An example: $x=(x_1,x_2,...)\in \ell^2({\Bbb N})$
$$ A x= (x_1,2x_2,3x_3,...),  \ \ \ (\lambda-A)^{-1} x = \left( \frac{1}{\lambda- k} x_k \right)_{k\geq 1}$$ 
which is readily verified to be compact for all $\lambda\notin {\Bbb N}$.
Another example (more interesting) is the Laplacian on a bounded domain $\Omega\subset \Bbb{R}^d$.