Operators whose spectrum has a finite number of connected component

Question

Assume that $H$ is a separable Hilbert space. Let $Q$ be the set of all operators$T \in B(H)$ such that the spectrum of $T$ has a finite number of connected component. Is $Q$ a subvector space of $B(H)$? What id $\bar{Q}$, the closure of $Q$? What is the closure of the span of $Q$?

Answer 1

Let $E=\{1,1/2,1/3,\cdots\}$ and let $\mu_{a}$ be the finite atomic meausre on $[0,1]$ that is supported on $E$ with $\mu\{1/n\}=1/n^{2}$. Let $\mu = \mu_{a}+m$ where $m$ is Lebesgue measure on $[0,1]$.

Let $X=L^{2}_{\mu}[0,1]$. Let $\chi$ be the characteristic function of $E$, and define operators $A, B \in \mathcal{L}(X)$ by $$ Af = xf(x),\;\;\; Bf =x\{1-\chi(x)\}f(x). $$ Then $\sigma(A)=[0,1]$ and $\sigma(B)=[0,1]$, but $\sigma(A-B)=E\cup\{0\}$. So you have two operators with connected spectrum and the sum of the two has spectrum consisting of infinitely many disjoint components.