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?
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That would be the boundary of $K$, wouldn't it? – Karolis Juodelė Feb 20 '13 at 19:24
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not necessarily – Emanuele Paolini Feb 20 '13 at 19:35
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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.
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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 $\mathbf S_2\setminus K$.
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