Spectrum of Element of Banach Algebra is Bounded

Theorem
Let $\struct {A, \norm {\, \cdot \,} }$ be a Banach algebra over $\C$.

Let $x \in A$.

Let $\map {\sigma_A} x$ be the spectrum of $x$ in $A$.

Then $\map {\sigma_A} x$ is bounded, and in particular:
 * $\cmod \lambda \le \norm x$ for all $\lambda \in \map {\sigma_A} x$

Proof
suppose that $A$ is unital, swapping $A$ for its unitization if necessary.

Let $\map G A$ be the group of units.

Let $\lambda \in \C$ be such that $\cmod \lambda > \norm x$.

Then from, we have:
 * $\ds \norm {\frac x \lambda} < 1$

From Element of Unital Banach Algebra Close to Identity is Invertible:
 * $\ds {\mathbf 1}_A - \frac x \lambda \in \map G A$

Hence:
 * $\lambda {\mathbf 1}_A - x \in \map G A$

So, we have:
 * $\set {\lambda \in \C : \cmod \lambda > \norm x} \subseteq \set {\lambda \in \C : \lambda {\mathbf 1}_A - x \in \map G A}$

so that:
 * $\set {\lambda \in \C : \lambda {\mathbf 1}_A - x \not \in \map G A} \subseteq \set {\lambda \in \C : \cmod \lambda \le \norm x}$

So we have:
 * $\map {\sigma_A} x \subseteq \set {\lambda \in \C : \cmod \lambda \le \norm x}$

as required.