User:Caliburn/s/fa/Spectrum of Linear Operator is Closed

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

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

Let $\map \sigma A$ be the spectrum of $A$.

Then $\map \sigma A$ is closed.

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

Then from the definition of the spectrum of $A$, we have:


 * $\map \sigma A = \C \setminus \map \rho A$

From Resolvent Set of Linear Operator is Open, we have:


 * $\map \rho A$ is open.

So, from the definition of a closed set, we have:


 * $\map \sigma A$ is closed.