Simply Connected Complex Domains

178 Views Asked by At

I am studying Complex Analysis using Sarason's text. In it he says that a domain $G \subset \mathbb{C}$ is simply connected if the extended complement $\bar{\mathbb{C}} \setminus G$ is a connected set. Throughout the text a domain is a nonempty, open connected set.

Is there a way to prove that convex domains are simply connected using just this definition? Sarason provides a handful of other notions that are equivalent to simple connectedness (namely the Winding Number Criterion) but I'm having trouble wrapping my head around how an extended complement can be regarded as a connected set.

1

There are 1 best solutions below

0
On BEST ANSWER

Let $ G \subseteq \mathbb{C} $ be a convex domain and take a point $ z \in \mathbb{C} \setminus G $. Since $ G $ is convex, either $ \{z + t \mid t \geq 0\} $ or $ \{z - t \mid t \leq 0\} $ is disjoint from $ G $. This ray connects $ z $ to the point of infinity in $ \bar{\mathbb{C}} $.