Proposition 29.4.8.label Let $A$ be a unital Banach algebra and $x \in A$, then $\sigma_{A}(x)$ is a compact subset of $B_{\complex}(0, \norm{x}_{A})$.
Proof. By Proposition 29.2.3, $G(A)$ and hence the resolvent set of $x$ is open, so $\sigma_{A}(x)$ is closed. By Proposition 29.4.7, $[x]_{sp}\le \norm{x}_{A}$.$\square$
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