Let $K$ be a compact subset of $\mathbb R^2$ such that $\mathbb R^2\setminus K$ is not connected. Is it true that $K$ contains a simple closed curve?
$\begingroup$ $\endgroup$
2 
- $\begingroup$ That would be the boundary of $K$, wouldn't it? $\endgroup$Karolis Juodelė– Karolis Juodelė2013-02-20 19:24:53 +00:00Commented Feb 20, 2013 at 19:24
- $\begingroup$ not necessarily $\endgroup$Emanuele Paolini– Emanuele Paolini2013-02-20 19:35:01 +00:00Commented Feb 20, 2013 at 19:35
Add a comment |
2 Answers
$\begingroup$ $\endgroup$
No. One counterexample is the closure of topologist's sine curve $y=\sin (1/x)$, $0<x<1$, plus a curve connecting its "good" end to the line segment at the other end. The complement is open and clearly not path-connected, hence not connected.
$\begingroup$ $\endgroup$
1 However, it is true that if $K$ is a compact subset of the plane, connected and locally connected, and non-empty, then $\mathbf R^2\setminus K$ is connected if and only if $K$ is simply connected, if and only if $K$ is contractible to a point, if and only if $K$ is a deformation retract of $\mathbf R^2$. The main difficulty is Caratheodory's theorem that shows that there exists a continuous map from the closed unit disk to $\mathbf S_2\setminus \mathring K$ which induces a homeomorphism from the open unit disk to $\mathbf S_2\setminus K$.
- $\begingroup$ Do you know any reference for this fact? $\endgroup$a.s. graduate student– a.s. graduate student2025-01-03 15:56:29 +00:00Commented Jan 3 at 15:56