Complement is connected iff Connected components are Simply Connected

Question

Let $G$ be an open subset of $\mathbb{C}$. Prove that $(\mathbb{C}\cup \{ \infty\})-G$ is connected if and only if every connected component of $G$ is simply connected.

Intuitively it holds, but I am stuck on both directions. For the straight one if we take a connected component of G, say H, and suppose that it is not simply connected , then $\overline C-H$ is not connected. Thus $\overline C-H$ is written as a union of two closed disjoint nonempty sets K and L. From this we should derive absurdity... — Dimitris, Mar 14 '13 at 17:50
Would this also hold for higher dimensions? If no, would there be some counterexample? — xxxg, Feb 14 '25 at 15:30
@xxxg: A torus is not simply connected but its complement in the $3$-dimensional space it is embedded in, is connected. If you really want an open subset, choose a open tubular neighbourhood of it. — Loulou, Mar 11 '25 at 21:37

score 5 · Accepted Answer · answered Mar 14 '13 at 18:16

5

If the connected open set $H \subset \mathbb C$ is not simply connected, there is a simple closed curve $C$ in $H$ that is not homotopic to a point in $H$. Therefore there must be points inside $C$ that is not in $H$. Such a point and $\infty$ are in different connected components of $({\mathbb C} \cup \{\infty\} - G$.

answered Mar 14 '13 at 18:16

Robert Israel

470,583

Thank you, but I cannot make it clear how exactly this is of help. – Dimitris Mar 14 '13 at 18:41
@Dimitris: He did one direction for you (by contrapositive). – Cameron Buie Mar 14 '13 at 22:37

Complement is connected iff Connected components are Simply Connected

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