generalization of the Jordan curve theorem

Question

Here is a plausible generalization of Jordan curve theorem which I couldn't find a rigorous proof for it.

Let $K$ be a compact subset of $\mathbb{R}^2$ which is homotopic equivalent to $S^1.$ Prove that $\mathbb{R}^2-K$ has two connected components, one is bounded while the other is not.

This should follow from a homology calculation (basically the same one as for the standard Jordan theorem). — Miha Habič, Jun 26 '11 at 08:28
More specifically, it follows from Alexander duality (http://en.wikipedia.org/wiki/Alexander_duality). — George Lowther, Jun 26 '11 at 17:24
@ All: My apologies, I misread the question. If possible, I can delete it. — gary, Jun 27 '11 at 20:19

gary · Answer 1 · 2011-06-26T17:56:52.303

2

This is true for $\mathbb R^2$, but not for dimensions 3-and-higher; the general issue is dealt with by Schoenflies. See:

http://en.wikipedia.org/wiki/Schoenflies_problem

This is related (maybe equivalent) to the fact that there are no knots in $\mathbb R$ nor in $\mathbb R^2$

edited Jun 26 '11 at 17:56

answered Jun 25 '11 at 23:57

gary

4,027

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But Alexander's Horned Sphere is not a counterexample to the question here. It is a counterexample to the statement that $\mathbb{R}^2\setminus K$ is has two simply connected components. – George Lowther Jun 26 '11 at 17:09
Right, my bad. I will edit it out. – gary Jun 26 '11 at 17:55

generalization of the Jordan curve theorem

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