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As a student, I was taught that the Jordan curve theorem is a great example of an intuitively clear statement which has no simple proof.

What is the simplest known proof today?

Is there an intuitive reason why a very simple proof is not possible?

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One reason why a simple proof is hard to come up with is that curves can be fiendishly complicated. In fact, if you restrict attention to piecewise smooth curves, it is not hard to come up with a simple proof, the point being that a smooth curve really divides the plane locally. –  Harald Hanche-Olsen Dec 11 '09 at 2:38
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In the smooth, PL or PL-smooth case the proof is quite intuitive and straightforward -- IMO going further to prove the Schoenflies theorem (that one of the bounded regions is a disc) is similarly straightforward. The reason it's not simple in the topological case is that topological curves can be extremely "fuzzy" making local arguments difficult -- Julia sets that are simple closed curves, for example. –  Ryan Budney Dec 11 '09 at 2:46
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For general continuous curves, it's not that a simple proof is not possible, it's that it's not desirable. The true content of the result is homology theory, which proves the separation result in n dimensions. There are special proofs in 2D that are simpler, but every such proof that I have seen feels like a one-night stand. –  Greg Kuperberg Dec 11 '09 at 5:34
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"One reason why a simple proof is hard to come up with is that curves can be fiendishly complicated." @Harald: I guess a general continuous function from R to R can be fiendishly complicated, and that's why we shouldn't expect a simple proof of the intermediate value theorem, right? ;-) –  Kevin Buzzard Dec 11 '09 at 7:46
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@buzzard: But there is a simple proof of the intermediate theorem. I guess that is what your smiley is about. The problem for the Jordan theorem is with showing, even locally, that the curve has two sides. For the intermediate value problem, the corresponding difficulty would be with proving that every function value is either positive, negative, or zero. @Konrad: So what? Part of the purpose of heavy machinery is precisely to enable slick proofs. Or proofs at all. (The deeper purpose is to provide insight.) –  Harald Hanche-Olsen Dec 11 '09 at 16:31

9 Answers 9

up vote 17 down vote accepted

There's a short proof (less than three pages) that uses Brouwer's fixed point theorem, available here:

The Jordan Curve Theorem via the Brouwer Fixed Point Theorem

The goal of the proof is to take Moise's "intuitive" proof and make it simpler/shorter. Not sure whether you'd consider it "nice," though.

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Thank you for this reference. The quality of the linked PDF is awful, however. For those with the requisite access, the paper is by Ryuji Maehara, Amer. Math. Monthly 91 (1984), 641–643 (doi:10.2307/2323369). The reviewer (see ams.org/mathscinet-getitem?mr=769530) also recommends the proof in Munkres' Topology: a first course as requiring a “comparable quantity of background”. –  Harald Hanche-Olsen Dec 11 '09 at 3:17
    
The least nice part of Maehara's proof is the dependence on the Tietze extension theorem. –  Konrad Swanepoel Dec 11 '09 at 19:36
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In my opinion the Tietze-trick is the most beautiful part of the proof. I am also one of the many people grown up having told that the Jordan curve theorem is something quite difficult to prove. But after reading this proof I will sleep very well. –  Balazs Strenner May 31 '11 at 4:47

It depends on what you mean by "simple". If you know homology, the proof is not very hard (less than 1 page), see for example, section 2.B ("Classical Applications") of Hatcher's book "Algebraic Topology".

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Several proofs are here:

http://www.maths.ed.ac.uk/~aar/jordan/index.htm

Among them, Tverberg's (1980) could (and should) be mentioned.

But, after reading (and reading)

http://www.math.sunysb.edu/~bishop/classes/math401.F09/HalesDefense.pdf ,

I really like Jordan's proof itself.

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It is "Tverberg", not "Tvelberg". Sorry for the typo. –  Ady Dec 11 '09 at 17:29
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I've fixed it for you. But you can edit your answers, you know. –  Kevin H. Lin Dec 26 '09 at 10:11
    
Well, nobody asked you to do that, I think. –  Ady Dec 27 '09 at 17:07
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Carsten Thomassen's proof is relatively simple:

Carsten Thomassen, The Jordan-Schönflies theorem and the classification of surfaces. Amer. Math. Monthly 99 (1992), no. 2, 116-130.

By the way, the Jordan Curve Theorem has a formal proof (one that can be checked by a computer): Thomas C. Hales, The Jordan curve theorem, formally and informally. Amer. Math. Monthly 114 (2007), no. 10, 882-894.

Hales bases the formal proof on Thomassen's.

The following is a survey on the older papers on the subject:

H. Guggenheimer, The Jordan curve theorem and an unpublished manuscript by Max Dehn. Archive for History of Exact Sciences 17 (1977), 193-200.

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Konrad: Do you understand what Thomassen argues for the Schonflies part of the theorem? It is the most complicated part of his paper. I admit that I did not try all that hard to follow it, but still I got lost. –  Greg Kuperberg Dec 11 '09 at 18:40
    
@Greg: No, I only looked at the first part. I have never grokked a proof of Jordan-Schönflies. –  Konrad Swanepoel Dec 11 '09 at 19:34

There is a proof of the Jordan Curve Theorem in my book "Topology and groupoids" which also derives results on the Phragmen-Brouwer Property. Also published as

`Groupoids, the Phragmen-Brouwer property and the Jordan curve theorem', J. Homotopy and Related Structures 1 (2006) 175-183.

The van Kampen Theorem for the fundamental groupoid on a set of base points is used to prove that if $X$ is pathconnected and the union of open path connected sets $U,V$ whose intersection has $n$ path components, then the fundamental group of $X$ contains the free group on $n-1$ generators as a retract.

May 30: The question asks why there is not a simple proof. Perhaps the following Figure 9.10 from the above book will explain why a proof is not expected to be so so easy; how do you decide whether a point in the middle is inside or outside?

Fig9.10

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You question seems to have many easy intuitive answers. For example, fix the point $x$ you want to decide is either inside or outside. Bound the given diagram by a large enough circle, and fix a point $y$ outside the circle. Connect $x$ and $y$ by a straight line. Now just count the number of times this line crosses the boundary of the figure. If it is odd, the point x lies inside. If it is even, the opposite. (Of course, this only works because your curve is piece-wise linear.) –  Pace Nielsen May 30 '12 at 19:18
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I kind of left it to the imagination to concoct a modified example which nullifies an immediate proposed method. The given figure is rectilinear; it might be modified to have infinitely many wobbles, and then the line chosen might cross the figure say countably infinitely many times. The other method for this example is of course to start filling in from a point near the outside edge. Just the job for a child! –  Ronnie Brown Jul 20 '12 at 20:05

You should compare with: "Geometric Topology in Dimensions 2 and 3", Moise, Edwin E. (1977). Springer-Verlag and tell

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An elementary proof by means of nonstandard analysis (by reduction to the case of polygons) and elementary combinatorics is given in Kanovei & Reeken, A nonstandard proof of the Jordan curve theorem, RAE 1999, 24, 161--170

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There's a remarkable elementary proof of the Jordan separation theorem, using only the fundamental group, due to Doyle. The proof is expounded in detail in Armstrong's book Basic Topology, Section 5.6.

I think this approach could be extended to prove that there are two complementary components. If there were more, then by an application of Van Kampen's theorem, one could conclude that the fundamental group is a free group of rank $>1$, which would give a contradiction as in Doyle's argument.

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A nice and simple proof using mod 2 intersection theory is given in the book Differential Topology by Guillemin,Pollack.

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Isn't this the <b>smooth</b> Jordan curve theorem? (Much easier). The function f (Page 86) is assumed to be smooth. –  Daniel Moskovich May 30 '12 at 13:08

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