# Proof of theorem of connectedness of an open set

I was watching an online lecture about complex analysis and in one if the first videos: The following theorem is stated:

Let $$G$$ be an open set in $$\mathbb{C}$$. Then $$G$$ is connected if and only if any two points in $$G$$ can be joined by successive line segments.

This theorem seems pretty elementary, but comes up time and time again throughout the lecture series. Does anyone know the proof of it?

The proof isn't that difficult. It follows because $$\Bbb C$$ (or more generally, $$\Bbb R^n$$) has a basis for the topology consisting of convex sets. In particular, the basis of all balls. Note that the condition that any two points of $$G$$ can be connected by a curve, a continuous map from $$[0,1]$$ into $$G$$, is called "path connectedness". Your condition is stronger, saying that the curves are polygonal (consisting of line segments strung together).
Path connectedness implies connected: Suppose $$G$$ is path connected, but has a disconnection by non-empty open sets $$A, B$$. Choose $$a \in A, b \in B$$. Then there is some curve $$\gamma: [0,1] \to G$$ such that $$\gamma(0) = a$$ and $$\gamma(1) = b$$. But then $$\gamma^{-1}(A)$$ and $$\gamma^{-1}(B)$$ would form a disconnection of $$[0,1]$$. Since $$[0,1]$$ is connected, no such disconnection exists. Therefore $$G$$ must be connected.
Conversely, suppose $$G$$ is connected, and let $$x \in G$$. And let $$A$$ be the set of all points of $$G$$ that can be connected to $$x$$ by a polygonal path in $$G$$. Let $$B = G\setminus A$$ be all the other points in $$G$$. For any $$a \in A$$, since $$G$$ is open, there is some ball about $$a$$ within $$G$$. But any point of that ball can be connected to $$a$$ by a line segment, and since $$a$$ has a polygonal path to $$x$$, so do all the other points in the ball. Therefore the entire ball must be in $$A$$. Thus, A must be open. Similarly any $$b \in B$$ also has a ball around it in $$G$$. If some point $$c$$ in the ball had a polygonal path back to $$x$$, then that path could be extended by the line segment from $$c$$ to $$b$$, contradicting that $$b \in B$$. Therefore no point in the ball can have a polygonal curve $$x$$, and the entire ball is in $$B$$. Thus $$B$$ is also open. If $$B$$ is non-empty, this would form a disconnection of $$G$$, which cannot be. Therefore $$B = \emptyset$$, and $$G = A$$.