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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?

Sam Nead
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user2498
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    I retagged this a bit; proof-theory is inappropriate here. – Harald Hanche-Olsen Dec 11 '09 at 02:35
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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 02:38
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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 02:46
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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 05:34
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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 07:46
  • This is also something that's been bothering me. On the one hand there are these slick proofs requiring heavy machinery, and on the other hand there are more pedestrian ones that go on for ages and are not particularly insightful. – Konrad Swanepoel Dec 11 '09 at 13:50
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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
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    @Harald: I don't have a problem with the heavy machinery in the slick proofs. My point is that I would like something inbetween the two extremes: almost slick and almost elementary. – Konrad Swanepoel Dec 11 '09 at 18:18
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    @Harald: yes, of course. You seemed to be saying "it will be hard to find a simple proof of a statement about curves, because curves can be very complicated". I was giving an example where I pushed this logic a little further and it seemed to me to break down. – Kevin Buzzard Dec 11 '09 at 21:40
  • Maybe you have noticed Thurston's answer here. It looks like the shortest proof of a related statement http://mathoverflow.net/questions/66048/riemann-mapping-theorem-for-homeomorphisms/66061#66061 – Dmitri Panov Jun 15 '11 at 21:21

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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.

TerronaBell
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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 http://www.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 03:17
  • 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 04:47
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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".

Kevin H. Lin
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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

Feb 9, 2016: A small correction is needed, and this is given in this paper jointly with Omar Antolin-Camarena.

October 26, 2016 Related issues on many base points are discussed in this paper.

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Ronnie Brown
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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
  • This is irrelevant for the purpose of rigorous proof, but I can tell you how software engineers "decide whether a point in the middle is inside or outside": virtually every piece of software I saw starts by assigning direction to the boundary segments, draws a random ray from the point of interest to outside of the polygon, and counts signed intersections of the ray with boundary segments. The sign of the intersection comes from the cross product of the ray with the boundary segment. – Michael Jun 12 '17 at 16:18
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    Just to add some relevant context: the illustration in this answer appears to be a variation of the fourth iterate of the Peano-Gosper-curve iteration (aka the flowsnake iteration). – Peter Heinig Aug 28 '17 at 18:52
  • @Michael: That method, applied to higher iterates of the Peano-Gosper curve, would quickly run into an exponential explosion, given that the number of individual boundary segments grows exponentially. – Lee Mosher Dec 28 '17 at 13:58
  • The curve in the picture looks in the opposite direction it should. After a convenient smoothing this curve is still as complicate as the original one but smooth. The deep difference is local. A continuous curve doesn’t simplify if you zoom in a point. Try with $x\cdot \sin\frac{1}{x}$ at the origin. – Claudio Rea Apr 09 '20 at 19:32
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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.

Kevin H. Lin
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Ady
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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.

Konrad Swanepoel
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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
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    Guggenheimer criticizes Jordan's original proof, but as mentioned in Ady's answer, Hales defends Jordan's proof. See also this comment. – Timothy Chow Aug 18 '20 at 18:43
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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

Vladimir Kanovei
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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.

Ian Agol
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  • Wow, this proof is amazing! Doyle's exposition leaves a lot to be desired, but the reference to Armstrong was super helpful. And your suggestion to strengthen the result was fun too, since I've never actually had a chance to apply Van Kampen for groupoids. – Nikhil Sahoo Feb 21 '22 at 15:32
  • I quite like how we're basically using the fact that any Jordan curve in $\mathbb R^2$ admits an ambient isotopy in $\mathbb R^3$ to the standard circle. This definitely prompts the impulse to apply Jordan-Schoenflies, but that extra dimension of wiggle room works its magic! – Nikhil Sahoo Feb 21 '22 at 15:44
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You should compare with: "Geometric Topology in Dimensions 2 and 3", Moise, Edwin E. (1977). Springer-Verlag and tell

janmarqz
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I'm sure there was a simple proof of the Jordan Curve Theorem as one of Kevin Brown's math articles, at http://www.mathpages.com.

As I recall, it was based on counting curve crossings over each of a sequence of concentric narrow annuli, the outmost of which entirely encloses the curve.

But as I can't now find it on Kevin's site, and there is a host of other fascinating articles there which will doubtless divert your attention for quite some time, I fear this reply probably isn't one of my more helpful ones!

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

Alexey Ustinov
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Marcus
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