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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10$\begingroup$ 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. $\endgroup$– Harald Hanche-OlsenCommented Dec 11, 2009 at 2:38
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4$\begingroup$ 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. $\endgroup$– Ryan BudneyCommented Dec 11, 2009 at 2:46
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37$\begingroup$ 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. $\endgroup$– Greg KuperbergCommented Dec 11, 2009 at 5:34
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14$\begingroup$ "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? ;-) $\endgroup$– Kevin BuzzardCommented Dec 11, 2009 at 7:46
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5$\begingroup$ @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. $\endgroup$– Kevin BuzzardCommented Dec 11, 2009 at 21:40
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10 Answers
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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.
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2$\begingroup$ 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”. $\endgroup$ Commented Dec 11, 2009 at 3:17
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$\begingroup$ The least nice part of Maehara's proof is the dependence on the Tietze extension theorem. $\endgroup$ Commented Dec 11, 2009 at 19:36
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5$\begingroup$ 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. $\endgroup$ Commented May 31, 2011 at 4:47
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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".
answered Dec 11, 2009 at 3:06
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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?
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.
answered Jan 8, 2012 at 10:25
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10$\begingroup$ 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.) $\endgroup$ Commented May 30, 2012 at 19:18
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4$\begingroup$ 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! $\endgroup$ Commented Jul 20, 2012 at 20:05
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$\begingroup$ 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. $\endgroup$– MichaelCommented Jun 12, 2017 at 16:18
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1$\begingroup$ 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). $\endgroup$ Commented Aug 28, 2017 at 18:52
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$\begingroup$ @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. $\endgroup$ Commented Dec 28, 2017 at 13:58
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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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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.
answered Dec 11, 2009 at 14:21
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$\begingroup$ 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. $\endgroup$ Commented Dec 11, 2009 at 18:40
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$\begingroup$ @Greg: No, I only looked at the first part. I have never grokked a proof of Jordan-Schönflies. $\endgroup$ Commented Dec 11, 2009 at 19:34
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1$\begingroup$ Guggenheimer criticizes Jordan's original proof, but as mentioned in Ady's answer, Hales defends Jordan's proof. See also this comment. $\endgroup$ Commented Aug 18, 2020 at 18:43
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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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$\begingroup$ 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. $\endgroup$ Commented Feb 21, 2022 at 15:32
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$\begingroup$ 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! $\endgroup$ Commented Feb 21, 2022 at 15:44
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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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You should compare with: "Geometric Topology in Dimensions 2 and 3", Moise, Edwin E. (1977). Springer-Verlag and tell
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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!
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$\begingroup$ Why don't you try to reconstruct the proof, to find out wether it actually works out? archive.org does not have it, either. $\endgroup$ Commented Feb 9, 2016 at 21:13
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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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3$\begingroup$ Isn't this the <b>smooth</b> Jordan curve theorem? (Much easier). The function f (Page 86) is assumed to be smooth. $\endgroup$ Commented May 30, 2012 at 13:08