Understanding connectedness argument in proof of Analytic Fredholm Theorem

Question

Let $X$ be a complex Banach space, and let $D \subset \mathbb{C}$ be a domain. Let $\mathcal{L}(X)$ denote the Banach space of bounded linear transformations $X \to X$.

The Analytic Fredholm Theorem is given as follows:

Let $A: D \to \mathcal{L}(X)$ be an operator-valued analytic function such that $A(z)$ is a compact operator for each $z \in D$. Then either:

1. $(I - A(z))^{-1}$ does not exist for any $z \in D$, or

2. $(I - A(z))^{-1}$ exists for all $z \in D \setminus S$, where $S$ is a discrete subset of $D$.

This is Theorem 8.26 from Inverse Acoustic and Electromagnetic Scattering Theory by Colton and Kress. To prove this theorem, the authors first establish that, for each $z_0 \in D$, there exists a ball $B_{r_{z_0}}(z_0) \subset D$ within which either 1. or 2. holds. Then, the authors claim that the theorem itself follows by a simple connectedness argument.

I am having trouble understanding how the connectedness argument should go.

Here's what I have so far. To prove the full theorem, assuming we know the local version, let's assume that 1. does not hold. Then we want to show that

$$ W =\{ z \in D | (I- A(z))^{-1} \text{ does not exist}\}$$

is discrete. If we fix $z_0 \in W$, then we know that in $B_{r_{z_0}}(z_0)$, either 1. or 2. holds. If we can rule out 1., we are done. But I'm not sure how to continue from here. I haven't really brought in the connectedness property of $D$, so I think I am missing a substantial step.

Hints or solutions are greatly appreciated.

Answer 1

The argument is

Every $z\in D$ has a neighbourhood $U$ such that either property 1 holds for all $w\in U$, or property 2 holds for all $w\in U$. Then, since $D$ is connected, either property 1 holds for all $w\in D$, or property 2 holds for all $w\in D$.

What the properties are is not relevant for the connectedness argument, they just need to be incompatible. The argument is completed by defining

$$A = \{ z \in D : \text{property 1 holds for all } w \text{ in a neighbourhood of} z\}$$

and

$$B = \{ z \in D : \text{property 2 holds for all } w \text{ in a neighbourhood of} z\}.$$

By construction, the sets $A$ and $B$ are open. By assumption, $D = A \cup B$. And since for no point both properties can hold together, we have $A\cap B = \varnothing$. Then the connectedness of $D$ implies that $A = \varnothing$ (hence $D = B$) or $B = \varnothing$ (hence $D = A$).

Here, property 1 is

There is a neighbourhood $U$ of $z$ such that $(I - A(w))^{-1}$ does not exist for any $w\in U$.

And property 2 is

There is a neighbourhood $U$ of $z$ such that $S = \{w\in U : (I-A(w))^{-1} \text{ does not exist}\}$ is discrete in $U$.