According to Wikipedia, in the complex plane, an open set, say $X$, is simply connected if both $X$ and its complement are connected on the Riemann sphere.

Consider the complement of the closed unit disk, this set is open, and both the complement and the set are connected (at least that's what I would assume); however, this set isn't simply connected.

My question, is it possible to construct the complement of the closed unit disk using two disjoint open sets, i.e., is $\{\infty\}$ considered an open set?

Help me to understand, using this definition, why the set is not simply connected, and perhaps the significance of using the Riemann sphere in defining simple connectedness.

  • $\begingroup$ That definition seems wrong. Can you link the Wikipedia page? $\endgroup$
    – Javier
    Feb 7 at 2:25
  • $\begingroup$ en.wikipedia.org/wiki/Simply_connected_space $\endgroup$
    – Seth
    Feb 7 at 3:38
  • $\begingroup$ @Javier 4th paragraph of "Definition and equivalent formulations" $\endgroup$
    – Seth
    Feb 7 at 3:41

The set is simply connected in the Riemann sphere. Any contour around the hole can be deformed continuously into a contour around the point at infinity and you can collapse the latter into a point (the point at infinity, say).

  • $\begingroup$ I assumed the set was not simply connected using a definition that roughly states that the complement must be 'connected to infinity'. $\endgroup$
    – Seth
    Feb 7 at 1:32
  • $\begingroup$ In that case, an annulus would be simply connected.. $\endgroup$
    – GReyes
    Feb 7 at 2:03
  • $\begingroup$ To me, simply connected means that any closed contour inside the region can be collapsed to a point. $\endgroup$
    – GReyes
    Feb 7 at 2:04
  • $\begingroup$ The definition I'm referring to assumes that any point in the complement can be connected to infinity. The 'inside' of the annulus, i.e, in the complement (not the interior of the annulus), cannot 'connect to infinity'. See Definition $8.1$ of Complex Analysis by Bak and Newman for the precise definition as there is the possibility that my understanding of it may be incorrect. $\endgroup$
    – Seth
    Feb 7 at 3:54

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