1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I see.. thanks! Evidently I need to study the notation of path-connectedness. (I had never seen it before and since it isn't discussed in the book, I was unaware of it.) $\endgroup$ – measuresproblem Apr 5 at 13:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.