# If $G$ is a convex set, prove that $\mathbb{C}\cup\{\infty\}\setminus G$ is connected.

Let $$G$$ be a convex set. I want to prove that $$G$$ is simply connected. This is my definition of simply connected:

A domain $$G$$ in $$\mathbb{C}$$ is said to be simply connected if $$\overline{\mathbb{C}}$$, its extended complement, is connected. A domain is a nonempty, connected, open subset of $$\mathbb{C}$$.

I know this problem has been posted on this site, but I haven't found on using this definition. This definition is from Sarason's complex analysis and I do not much (or, really, any) topology to follow the other answers.

I know this problem must be easy (the author asserts the the reader will easily verify this), but I really have no idea on how to start.

I thought about the particular case where $$G$$ is the unit disk. Then if I imagine $$G$$ on the Riemann sphere, it's not clear to me why the complement of $$G$$ cannot be written as the disjoint union of open sets.

We may assume without loss of generality that $$G$$ is properly contained in $$\mathbb C$$. Now let $$z$$ be in $$\mathbb C\setminus G$$ and consider an arbitrary line in $$\mathbb C$$ passing through $$z$$. This line is the union of two rays starting at $$z$$. I claim that at least one of those rays is disjoint from $$G$$. Indeed, if both contain a point of $$G$$, then by convexity $$z$$ is contained in $$G$$, contradicting the choice of $$z$$. Now any such ray gives us a path from $$z$$ to $$\infty$$ on the Riemann sphere, hence $$\mathbb C\cup \{\infty\}\setminus G$$ is path-connected and therefore connected.