Proposition 6.2.12.label Let $A$ be a unital Banach algebra and $U\suf \com$ open then $E_{U}=\curl{x\in A:\spec{A}{x}\suf U}\suf A$ is open.
Proof. Let $x_{0}\in E_{U}$, the function $\lam\in U^{c}\mto (x_{0}-\lam)\inv\in A$ is continuous with
\begin{align*}\limit{\abs{\lam}\to\infty}\norm{(x_0-\lam)\inv}{}=0\implies \SUP{\lam\in U^c}\norm{(x_0-\lam)\inv}{}<\infty\end{align*}
Let $\de=\INF{\lam\in U^c}\frac{1}{\norm{(x_0-\lam)\inv}{}}>0$. If $\norm{x-x_0}{}<\de$ then for any $\lam\in U^{c}$ we have
\begin{align*}\norm{(x_0-\lam)-(x-\lam)}{}<\frac{1}{\norm{(x_0-\lam)\inv}{}}\end{align*}
hence $(x-\lam)\inv$ exists by 6.1.17, so $\spec{A}{x}\suf U$ and $x\in E_{U}$ as desired.$\square$