Proposition 6.2.5.label Let $A$ be a Banach algebra then for any $x\in A$
Proof. By Lemma 6.1.7 we assume that $A$ is unital. Let $x\in A$ and consider the continuous map $f:\lam\in \com\mto x-\lam \in A$. The resolvent of $x$ is the inverse image $f\inv(G(A))$ thus is open by Proposition 6.1.17 so $\spec{A}{x}$ is closed. Suppose now $\abs{\lam}>\norm{x}{}$ then $\norm{\frac{x}{\lam}}{}<1$ so by Proposition 6.1.16 the series $\sums{\infty}{n=0}\braks{\frac{x}{\lam}}^{n}$ converges to $\braks{\frac{x}{\lam}-1}\inv$. Hence
\begin{align*}(x-\lam 1)\inv=\frac{1}{\lam}\parens{\frac{1}{\lam}-x}\inv\end{align*}
and $\lam$ is in the resolvent of $x$. Therefore, $\spec{A}{x}$ is contained in the disk $\abs{\lam}\leq \norm{x}{}$. Hence it is compact.$\square$