I am reading through the following statement and proof in Aupetit’s A Primer on Spectral Theory
He provides the following proof:
Towards the end of the proof, Aupetit says that the set $E$ as defined in the proof is both open and closed. I understand why the set is open, but I am struggling to see why it is closed. Can anyone please guide me in the right direction as to how I can go about showing $E$ is closed (so as to convince myself why it is true)?
It was shown that $\operatorname{Sp} f(\lambda)$ is locally constant on $D$; i.e. every point of $D$ has a neighborhood on which $\operatorname{Sp} f(\lambda)$ is constant. So for each set $A \subseteq \mathbb{R}$, the set $D_A = \{ \lambda \in D : \operatorname{Sp} f(\lambda) = A\}$ is open in $D$. Thus $E^c = \bigcup_{A \ne \operatorname{Sp} f(\lambda_0)} D_A$ is open, since it is a union of open sets.
There's really nothing about spectra or Banach algebras involved here. The argument just shows that if $X$ is any connected topological space and $S$ is any set, then any locally constant function $f : X \to S$ is constant. Here you may take $S = 2^{\mathbb{R}}$ if you like.