Spectrum of Element of Banach Algebra is Closed

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 closed.

Proof

Without loss of generality suppose that $A$ is unital, swapping $A$ for its unitization if necessary.

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

From Resolvent Set of Element of Banach Algebra is Open, $\map {\rho_A} x$ is open.

From the definition of the spectrum, we have $\map {\sigma_A} x = \C \setminus \map {\rho_A} x$.

From the definition of a closed set, $\map {\sigma_A} x$ is therefore closed.

$\blacksquare$