Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
This should follow from a homology calculation (basically the same one as for the standard Jordan theorem). –  Miha Habič Jun 26 '11 at 8:28
More specifically, it follows from Alexander duality (en.wikipedia.org/wiki/Alexander_duality). –  George Lowther Jun 26 '11 at 17:24
@ George Lowther: Thanks. That's it. –  Ehsan M. Kermani Jun 27 '11 at 10:22
@ All: My apologies, I misread the question. If possible, I can delete it. –  gary Jun 27 '11 at 20:19

1 Answer 1

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


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

share|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.