Suppose $\sigma(T)=\{e^{\frac{i}{n}}:n\in\mathbb{N}\}\cup\{0,1\}$. How can I deduce an appropriate Banach space $X$ and compact operator $T$ such that this is true?
Obviously, $\lambda\in\sigma(T)\iff \mathscr{R}(T-\lambda I)\ne X\text{ or }T-\lambda I$ is not injective.
Since $\sigma_{p}(T)\subset\sigma (T)$, where $\sigma_{p}(T)=\{\lambda\in\mathbb{C}:T-\lambda I\text{ is not injective }\}$, i.e. the set of eigenvalues, and each $x\in\mathscr{N}(T-\lambda I)$ must satisfy the equation $$Tx=\lambda x.$$
Clearly $0$ and $1$ cannot be in the same point spectrum. However, this does not rule them out of being in the spectrum. Ultimately, I was wondering if someone had an effective method for determining such $X$ and $T$?
$X=l^2$ and $T$ the diagonal matrix with diagonal entries from $\sigma (T)$