Let $T\in B(l^2)$ be s.t. $Tx=(\alpha_1 x_1, \alpha_2 x_2, \cdots )$, where the set of all $\alpha$ is dense in $[0,1]$.
I've shown that the set of all eigenvalues is $A=(\alpha_j)_1^\infty$. The resolvend operator, where it exists, is bounded. Therefore, the continuous spectrum is empty. The range of $T-\lambda I$ is the entire $l^2$. Therefore, the residual spectrum is empty.
But is this true? I never used the fact that $\alpha$ is dense in $[0,1]$, which leads me to believe that at least one of my conclusions above is false.
(If possible, I prefer hints over solutions.)
But the sets are not bounded for all $\lambda\notin A$!