Spectrum of Bounded Linear Operator is Closed

Theorem

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

Let $T$ be a bounded linear operator on $X$.

Then the spectrum $\map \sigma T$ of $T$ is a closed set in $\C$.

Proof

From Resolvent Set of Bounded Linear Operator is Open, the resolvent set $\map \rho T$ is open.

From the definition of spectrum, we have $\map \sigma T = \C \setminus \map \rho T$.

From the definition of a closed set, $\map \sigma T$ is closed set in $\C$.

$\blacksquare$