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Recently I have read the proof of the Jordan Curve Theorem in Munkres' Topology, I wonder whether there are some generalizations and corollaries on this theorem as follows:

  1. I know any simple closed curve separate $\mathbb R^2$ into two components, one is bounded and the other one is unbounded. How to prove that the bounded component is simply connected and the unbounded component is homeomorphic to $\mathbb R^2-\{0\}$?

  2. Can the theorem be generalized into higher dimension? Given a simple closed curve in $\mathbb R^n$, does there exist $U\in\mathbb R^n$ such that $U$ is a simply connected 2-manifold whose boundary is exactly the curve?

I believe the first question would be a simple consequence of Jordan Curve Theorem (although I cannot figure out a proof, can any one give me one if possible?). However I am not quite sure whether my second question is even related to the JCT.

Y.H. Chan
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  • For #1, you want the Jordan–Schoenflies theorem. – Chris Culter Dec 21 '15 at 04:49
  • Some references related to the question are given in this paper http://arxiv.org/abs/1404.0556 which also applies the algebra of groupoids rather than covering spaces as in Munkres' proof. Wilder's 1949 volume on "Topology of Manifolds" has lots of related results. – Ronnie Brown Dec 21 '15 at 11:56

2 Answers2

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The answer to 2 as stated is no. Any open set $U\subset\mathbb{R}^n$ will have to be an n-manifold. (This is not hard to prove at all). So the correct generalization will be with $n-1$ manifolds instead of curves. This is known as the Jordan Separation Theorem. However, the interior need not be homeomorphic to a sphere in higher dimensions. Check out the Alexander Horned Sphere, which is a counter example when $n=3$.

boxotimbits
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  • I have heard a theorem, it is stated that if $F$ is a vector field in $\mathbb R^3$ and $curl(F)=0$, then the work done along any simple closed curve is zero. The proof is that any simple closed curve in $\mathbb R^3$ can be written as a boundary of a simply connected surface, and the result follows from Stroke's theorem. Then is #2 correct in the case of $\mathbb R^3$? – Y.H. Chan Dec 21 '15 at 04:57
  • For the Alexander Sphere, one of the component is not simply connected while another one is. Is my statement true if I need only one component simply connected? – Y.H. Chan Dec 21 '15 at 05:00
  • A surface in $\mathbb{R}^3$ will not be an open subset. I do believe that any smooth simple closed curve in $\mathbb{R}^3$ is the boundary of some smooth surface. – boxotimbits Dec 21 '15 at 05:03
  • My mistakes, I forgot that a surface in $\mathbb R^3$ is not open. Let me remove this condition. – Y.H. Chan Dec 21 '15 at 05:06
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The entirety of knot theory is founded on the premise that the answer to your second question, in the case of $\mathbb{R}^3$, is "no". The first counterexample usually encountered is the trefoil knot.

Lee Mosher
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