Spectrum of Element of Banach Algebra is Compact

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 compact.

Proof

The result follows immediately from:

Spectrum of Element of Banach Algebra is Bounded

Spectrum of Element of Banach Algebra is Closed

$\blacksquare$