Resolvent Set of Element of Banach Algebra is Open

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

Let $x \in A$.

Let $\map {\rho_A} x$ be the resolvent set of $x$ in $A$.

Then $\map {\rho_A} x$ is open.

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

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

Define $R : \C \to A$ by:
 * $\map R \lambda = \lambda {\mathbf 1}_A - x$

We show that $R$ is continuous.

We have, for $\lambda, \mu \in \C$:

Let $\epsilon > 0$ and $\lambda, \mu \in \C$ be such that:
 * $\cmod {\lambda - \mu} < \epsilon$

Then, we have:
 * $\norm {\map R \lambda - \map R \mu} < \epsilon$

Hence $R$ is continuous.

From Group of Units in Unital Banach Algebra is Open, $\map G A$ is open.

From the definition of the resolvent set, we have:
 * $\map {\rho_A} x = \set {\lambda \in \C : \lambda {\mathbf 1}_A - x \in \map G A}$

That is:
 * $\map {\rho_A} x = R^{-1} \sqbrk {\map G A}$

Since $R$ is continuous, and $\map G A$ is open, $R^{-1} \sqbrk {\map G A}$ is open.

We conclude that $\map {\rho_A} x$ is open.