Proof, [Proposition I.2.9, Tak01]. Let $x \in A$ with $\sigma_{A}(x) \subset U$, and $\lambda \in U^{c}$. By Proposition 29.2.3, for any $y \in A$ with $\norm{y}_{A} \le \normn{(\lambda - x)^{-1}}_{A}^{-1}$, $\lambda - x - y \in G(A)$ as well. Since the mapping $\lambda \mapsto \normn{(\lambda - x)^{-1}}_{A}$ vanishes at infinity and $U^{c}$ is closed, $\delta = \inf_{\lambda \in U^c}\normn{(\lambda - x)^{-1}}_{A}^{-1}> 0$. Therefore for every $z \in B_{A}(x,\delta)$, $\sigma_{A}(x) \subset U$.$\square$