properties of the resolvent

Question

Let $A$ be an operator with compact resolvent in a Hilbert space. I want to ask if the resolvent set of this operator is a dence subset in $\mathbb{C}$?

Answer 1

Let $A$ be a closed linear operator (otherwise the question makes little sense) and $z_0\in \rho(A)$. From the identity $$ A-z=((A-z_0)^{-1}-(z-z_0)^{-1})(A-z_0)(z_0-z) $$ one can easily see that $z\in \sigma(A)$ if and only if $(z-z_0)^{-1}\in \sigma((A-z_0)^{-1})$.

If $(A-z)^{-1}$ is compact, then $\sigma((A-z)^{-1})$ has no accumulation point except $0$. Thus, $\sigma(A)$ has no accumulation point, so it certainly does not contain an open disk. It follows that $\rho(A)=\mathbb{C}\setminus\sigma(A)$ is dense in $\mathbb{C}$.