In the proof of theorem 2.1.11 from C*-Algebras and Operator Theory, by Gerald Murphy the author is trying to prove $\sigma_B(u) = \sigma_A(u)$, where $B$ is a C*-subalgebra of $A$ and $u \in B$ is a unitary element.
The author starts the proof by assuming $B$ has a hermitian element $b$. He mentions that since $\sigma_A(b) \subset \mathbb{R}$, it has no holes. I understand why $\sigma_A(b) \subset \mathbb{R}$, but don't understand why it has no holes in $\mathbb{R}$.
After example 1.2.6, the author defines 'holes' of $K \subset \mathbb{C}$ as connected components of $\mathbb{C} \setminus K$.
Since $\sigma_A(b) \subset \mathbb{R}$, the statement that $\sigma_A(b)$ has no holes would imply $\sigma_A(b)$ is connected!?
The book in question defines "holes" as bounded connected components of $\mathbb C\setminus K$. So the claim is that for compact $K\subset \mathbb R$, the set $\mathbb C\setminus K$ has no bounded connected components (it would of course be non-sensical to claim that it has no connected components at all). As noted in the comments, it is crucial that we look at the complement in $\mathbb R$ and not in $\mathbb C$.
In fact, $\mathbb C\setminus K$ is always path connected in this situation. To see this, it suffices to prove that every $\lambda\in (\mathbb C\setminus K)\cap \mathbb R$ can be connected by a path in $\mathbb C\setminus K$ with every element $\mu$ of $\mathbb C\setminus \mathbb R$ - connecting arbitrary pairs of elements in $\mathbb C\setminus K$ can then be achieved by concatenating paths.
The piecewise linear curve $$ \gamma\colon [0,1]\to\mathbb C,\,\gamma(t)=\begin{cases}\lambda+2it{\operatorname{Im \mu}}&\text{if }0\leq t\leq 1/2\\ (2t-1)\mu+(2-2t)(\lambda+i{\operatorname{Im \mu}})&\text{if }1/2<t\leq 1\end{cases} $$ connects $\lambda$ and $\mu$. Moreover, $\gamma$ only intersects $\mathbb R$ at $t=0$ and $\gamma(0)=\lambda\in \mathbb C\setminus K$. Thus $\operatorname{im}\gamma$ lies inside $\mathbb C\setminus K$ as claimed.