User:Caliburn/s/fa/Resolvent Set of Linear Operator is Open

Theorem
Let $\struct {X, \norm \cdot_X}$ be a Banach space over $\C$.

Let $A : X \to X$ be a linear operator.

Let $\map \rho A$ be the resolvent set of $A$.

Then $\map \rho A$ is open.

Proof
Let $\lambda \in \map \rho A$.

By the definition of an open set, we aim to show that there exists some real number $\delta > 0$ such that for all $\alpha \in \C$ with:


 * $\cmod {\lambda - \alpha} < \delta$

we have:


 * $\alpha \in \map \rho A$

That is, writing $\beta = \alpha - \lambda$, we want to find some $\delta > 0$ such that whenever:


 * $\cmod \beta < \delta$

we have:


 * $\lambda + \beta \in \map \rho A$