Complement of spectrum of a invertible self-adjoint element has no bounded components

Question

Let $A$ be a unital $C^*$-subalgebra of a $C^*$-algebra $B$ and $x$ be an invertible element of $B$. Let $y$ be an element of $B$ defined by $y=x^*x$. Now $x$ is invertible implies $y$ is invertible and $ 0\notin \sigma_B(y)$. Also it is obvious that $y$ is self-adjoint. Now we know that spectrum of $y$ in $B$ contained in $\mathbb R,$ that is, $\sigma_B(y)\subset \mathbb R.$ $''$So the complement of $\sigma_B(y)$ has no bounded components.$''$
I am not able to understand the last statement inside $''...''$, that is $''$So the complement of $\sigma_B(y)$ has no bounded components.$''$ I know $\sigma_B(y) \subset \mathbb R$ is compact and $|\lambda| \leq \|y\|$ for all $\lambda \in \sigma_B(y).$ Also since $y$ is self-adjoint we have: $$r(y) = \sup \{|\lambda|: \lambda \in \sigma_B(y)\}=\|y\|.$$ From here how can I say that complement of $\sigma_B(y)$ has no bounded components$?$ There may be a bounded connected component, for example if $\|y\|=1$ and $\sigma_B(y)=[-1,-1/2]\cup[1/2,1]$. Then $(-1/2,1/2)$ is a bounded component of complement of $\sigma_B(y)$. It may be very simple and I am so stupid for not to figure it out. Can you please help me to solve this. Thank you.