Apparently, if $A$ is a $C^\ast$-algebra and the spectrum $\sigma(a) \subseteq \mathbb R$ then $\sigma (a)$ does not have holes.
I read this in Murphy and then tried to prove it since I didn't believe it. But I can't seem to prove it (even though -- since it's mentioned without proof -- it must be trivial).
Can someone please explain to me how to prove that a real spectrum is simply connected?
Edit
This is from page 41:
Your statement "a real spectrum is simply connected" is wrong. To give a concrete example, let $A = B(\mathbb C^2)$, where $\mathbb C^2$ carries the standard inner product. Let $$ a = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} $$ Then $\sigma(a) = \{0,1\}$, which is not connected. Note that - in general - on $B(\mathbb C^n)$, the spectrum consists of at most $n$ points and is almost never connected.
As Murphy writes on page 11, proof of Thm. 1.2.8, by $\sigma_A(b)$ "having no holes", he means that $\mathbb C \setminus \sigma_A(b)$ is connected. But this is obviously true for bounded subsets of $\mathbb R$.