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Let $C$ be a simple (Jordan) polygon in the plane. I would like to prove the Jordan curve theorem for $C$ using an elementary method. I think that we can prove it using the winding number (via complex analysis) with respect to $C$. Am I mistaken?

Related question: Triangle inside a simply connected open subset of the complex plane.

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Makoto Kato
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  • Perhaps begin by triangulating the polygon. – Greg Martin Jul 04 '16 at 00:18
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    @GregMartin: It's non-trivial to prove that any polygon can be triangulated. Do not forget that you cannot assume the 'common knowledge' about the sum of interior angles without already knowing that the polygon has an interior! – user21820 Jul 27 '16 at 15:33
  • @user21820 I'm trying to prove it my way. Not yet successful. – Makoto Kato Jul 28 '16 at 08:43
  • @user21820 There's a thread called "Proof of Jordan curve theorem for polygons" in this site. The answer refers to a paper. Is your method similar to that of the paper? – Makoto Kato Jul 28 '16 at 09:05
  • @MakotoKato: I don't know why you always avoid giving the direct link. You refer to http://math.stackexchange.com/a/1452171. No my method is not the same, but the easiest way I can fix the hand-wavy part of their argument is via the ideas in my proof. The rest is quite different because my proof directly establishes the winding number and not just the existence of interior and exterior. Please also fix your own question to link to your previous question, instead of making people search for it. – user21820 Jul 28 '16 at 12:22
  • @user21820 I'm writing with my iPhone. I can't use my PC right now for some reason. I don't know how to write a direct link with my iPhone. – Makoto Kato Jul 28 '16 at 14:12
  • @MakotoKato: Hmm I'm surprised if you can't copy the address or click the share button. Anyway I'll edit your question for you then. – user21820 Jul 28 '16 at 15:08
  • @user21820 Thanks. Yes, I would like to know the sketch of your proof. – Makoto Kato Jul 28 '16 at 22:33
  • Guess what? I found a subtle flaw in the part of my proof that goes from turning number to interior and exterior. So all I have established is that the turning number is $1$ or $-1$, but I somehow can't easily get that there is an interior and exterior. However, I think I found a different proof based on modifying the triangulation proof, but don't feel like writing it up now... – user21820 Jul 29 '16 at 16:18
  • I edited your questions to make them easier for subsequent people to find. I hope you don't mind! – user21820 Aug 01 '16 at 06:35
  • @user21820 Thanks! – Makoto Kato Aug 01 '16 at 06:39

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Indeed you can use the winding number defined by the contour integral to define interior and exterior of a piecewise smooth non-self-intersecting closed curve in the complex plane. However, that is not what I would call elementary, and you still need powerful theorems about winding number that probably boil down to the Jordan curve for a polygon... Basically, how do you know that the winding number is always $0$ or $1$ or $-1$? To prove that it cannot be any other integer is the intrinsic core of the Jordan curve theorem.

See this post for an elementary proof of the Jordan curve theorem for polygons. We can now easily define the winding number of a polygon around a point in the following way. If the point is outside the polygon, the winding number is $0$. If the point is inside the polygon, the winding number is the sum of the turning angles, which can be easily shown to be either $1$ or $-1$ full turns since the sum of the internal angles is $(n-2)180^\circ$.

By the way, there is an alternative definition of winding number that does not involve complex analysis and works for arbitrary continuous curves, which I'll sketch out below.

Let $C$ be the curve in question and $P$ be a point not on $C$. By compactness $C$ is contained in a finite set of open disks that do not contain $P$. Then we can partition $C$ into finitely many sections each of which is contained in one of those disks, and define the winding number as the winding number of the polygon given by the points between sections, which is just the sum of the directed angles made by the polygonal sides with respect to $P$.

We can then prove that winding number changes continuously as $P$ moves continuously without touching $C$, and so path-connected regions have the same winding number. From the polygonal case we know that there are only two possible winding numbers ignoring sign, which means that the general rectifiable curve $C$ also divides the plane into two path-connected regions.

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