How does $E$ closed follow from the upper semicontinuity of the spectrum?

Question

Let $f$ be an analytic function for a domain $D$ of $\mathbb{C}$ into a Banach algebra $A$. Suppose that, for all $\lambda \in D$, $\text{Sp}f(\lambda)$ is finite or a sequence converging to $0$. Suppose that $\mu \neq 0$ and $\mu \in \text{Sp}f(\lambda_0)$ for some $\lambda_0 \in D$.

Consider the set $E = \{ \lambda \in D: \mu \in \text{Sp}f(\lambda) \}$.

B. Aupetit mentions in a proof he writes for Theorem 3.4.26 in his book A Primer on Spectral Theory, that this set $E$ is closed by the upper semicontinuity of the spectrum. I am struggling to see why this is true.

Can anyone please point me in the right direction as to how I can show that $E$ is closed by the upper semicontinuity of the spectrum?

Answer 1

Upper semicontinuity of the spectrum is the following statement: if $U$ is a neighbourhood of $\text{Sp}(x)$, then there is a neighbourhood $V$ of $x$ such that $\text{Sp}(y) \subset U$ for all $y \in V$.

Suppose $\lambda \in D \backslash E$. That is, $\lambda \in D$ but $\mu \notin \text{Sp}(f(\lambda))$. Thus $U = \mathbb C \backslash \{\mu\}$ is a neighbourhood of $\text{Sp}(f(\lambda))$. By upper semicontinuity of the spectrum, there is a neighbourhood $V$ of $f(\lambda)$ such that $\mu \notin \text{Sp}(y)$ for all $y \in V$. Since $f$ is continuous, $f^{-1}(V)$ is a neighbourhood of $\lambda$, and this is disjoint from $E$. Thus $D \backslash E$ is open.