I am horribly confused about Jordan's Curve Theorem (henceforth JCT). Could you give me some reason why should the validity of this theorem be in doubt? I mean for anyone who trusts the eye theorem is obvious. Therefore answers like "do not trust the eye" is not going to help me.

I am looking for answers along the following lines. If the JCT were true for the "obvious" reason, then it might lead to some contradiction someplace else. To be more concrete, I can give an analogy with another theorem for which I had similar feelings -- namely the unique factorization theorem for natural numbers which subsided when I learnt about Kummer Primes.

In case you think that I am being too demanding when I ask this question, here is another direction you could help me with. In that case, I would just like one or two quick sentences about your personal experience with Jordan's curve theorem -- kind of like when you had your aha moment with this theorem. Something like, "I see, now I know (or can guess) why proving it was such a big deal". Please reply when you get time -- I am horribly confused.

Thanks for your patience,

J. M. ain't a mathematician
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Akash Kumar
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  • While this is far from an answer, I would think that this is exactly the same as the Banach-Tarski paradox is a "paradox", despite the fact that the proof is not *that* complicated. – Asaf Karagila Sep 04 '11 at 16:56
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    @Akash: What are the "obvious reasons" for its truth? – Ryan Budney Sep 04 '11 at 17:09
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    I'm quite sure this doesn't count as graph theory, and I don't really see how this is a ‘big picture’ question. As for the question itself: the mind's eye is of very limited imagination; the existence of continuous space-filling curves should be enough to give pause to the thought that the Jordan curve theorem is obvious. It is certainly obvious in the cases we can easily imagine (e.g. piecewise linear curves) but the point is that our imaginations are woefully inadequate! – Zhen Lin Sep 04 '11 at 17:10
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    For example, do these obvious reasons also apply to embeddings of $S^2$ in $\mathbb R^3$, why or why not? Similarly, $S^3$ in $\mathbb R^4$? – Ryan Budney Sep 04 '11 at 17:11
  • @Zhen: I tried to find some more appropriate tags. – Ryan Budney Sep 04 '11 at 17:13
  • Ryan may be thinking of examples like the Alexander Horned Sphere which defy naive intuitions about what is possible. But a straightforward answer to the question would be that human intuitions about Jordan Curves, and what a typical curve might look like, are generally hopelessly inadequate - that's where the space-filling curves come in. – Mark Bennet Sep 04 '11 at 17:25
  • Space filling curves are never simple, though :) – Mariano Suárez-Álvarez Sep 04 '11 at 17:27
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    Even for polygons, the Jordan curve theorem is not that easy to prove in detail. See *What Is Mathematics?* by Courant and Robbins. – lhf Sep 04 '11 at 18:34
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    An interesting article is [Jordan’s Proof of the Jordan Curve Theorem](http://mizar.org/trybulec65/4.pdf) by Thomas C. Hales. – lhf Sep 04 '11 at 18:38
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    That lots of "obvious" things are false may be seen by looking at the book _Counterexamples in Analysis_. – Michael Hardy Sep 04 '11 at 21:21
  • Sorry for the late response. But here are a few comments that I would like to address. @ryan - my mathematical training has many gaps (i do not know real analysis, complex analysis and topology) and have a hard time grappling with those notions. As for the embeddings of $S^2$ in $R^3$ and so on, i would think about some more. I did not try hard to think about it, and so far from a far-off standpoint, the theorem "looks obvious". – Akash Kumar Sep 05 '11 at 00:38
  • @zhen and ryan - thanks for editing the tags. i encountered the topic of JCT in Deistel's text on graph theory and thats why i kept the tags as such. mark, lhf, michael, mariano - thanks for your inputs, i will read them up – Akash Kumar Sep 05 '11 at 00:41
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    This [MO-thread](http://mathoverflow.net/questions/8521/) contains links to various expositions of the Jordan curve theorem (among them the one by Hales that @lhf mentions). – t.b. Sep 05 '11 at 09:44

4 Answers4


There is exactly one way in which one can convince oneself that a statement is not obvious: try to prove it and look at your attempts very, very critically.

If you think you can come up with a proof of the curve theorem, edit it into the answer and we can help you dissect it :)

Later. Asaf observes that it may be the case that you are refering to "intuitively obviousness". Well... I tend to think that when someone says something is intuitively obvious without having a specific proof in mind, he is just waving hands in words. But there are two observations one can make which are independent of that.

First, the full Jordan curve theorem deals with arbitrary closed curves, and here ẗhe word "arbitrary" includes things that one usually does not think about, curves so complicated that one cannot make accurate pictures of them, so it is rather unlikely one has any intuition about them at all (at least, when encountering the theorem for the first time) This is a situation that comes all the time: one thinks a statement is intuitively true only because one is not familiar with the cases where it is not clearly true at all. One's intuition is built upon our experience, and since our experience is, by definition, limited, our intuition is limited, too.

In any case, I would suggest you try to prove the version of the Jordan curve theorem which deals with piecewise linear curves, that is, with polygonal closed curves (with finitely many segments). In this more restricted situation, we have eliminated all the wildness continuous arcs can have and are left with a geometrically sensible situation. But! It is nonetheless quite not obvious how to prove the theorem in this simple situation either, as you will find out when you try. (This version can be proved without the machinery used to prove the general theorem, though)

Mariano Suárez-Álvarez
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  • Mariano: I think that the OP means intuitively obvious. – Asaf Karagila Sep 04 '11 at 17:05
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    @Asaf: what do you mean by intuitively obvious? I think it helps to be specific about what intuitive mechanisms you're using. – Ryan Budney Sep 04 '11 at 17:10
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    Additionally, one person's "intuitive" may very well be another person's "murky"... – J. M. ain't a mathematician Sep 04 '11 at 17:20
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    @Ryan: I mean that when you draw a closed and simple curve it is obvious that the theorem is true. I'm not implying that the theorem is very obvious in the mathematical sense of the word. However it is not a theorem which comes as shocking as Banach-Tarski, for example. The Jordan Curve Theorem is only shocking when you realize it is not a simple proof. – Asaf Karagila Sep 04 '11 at 19:29
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    @Asaf: what intuition do you have / are you using? This is my 2nd request for clarification, and your previous response does not address my question. If you have intuition you should be able to explain it, and you haven't. A problem with drawing the curve yourself is you can be fooling yourself -- you've fixed the game, so you could say you've *designed* the curve to be easy for you. That's why I suggested in my post to get someone else to draw the curve for you. – Ryan Budney Sep 04 '11 at 19:47
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    @Ryan: Intuition does not mean that it can be explained. Intuition means that "this seems like a true statement." – Asaf Karagila Sep 04 '11 at 19:48
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    @Asaf: Then why does it seem like a true statement to you? You haven't described why. A skeptic might assume you've drawn a couple curves in the plane and then just assumed the general case rather than thinking about it. Giving up isn't intuition. – Ryan Budney Sep 04 '11 at 19:57
  • @Ryan: This seems like a dead end conversation, so I am not going to try any harder than this: For **me** intuition means the ability to "feel" somehow that this is the right direction. Sometimes it is a misguided feel, and then you have the joy of finding out that you were wrong after much work. This is not longer the days of Euler when one could work only by intuition. However, letting myself be *guided* by intuition, I can tell that this is a **very reasonable** theorem. Again, unlike Banach-Tarski. If your intuition is poor it means nothing about your mathematical ability, so worry not. – Asaf Karagila Sep 04 '11 at 20:00
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    @Asaf, but does that same intuition break somehow in higher dimensions? Do you *feel* the bounded component of a 2-sphere should be a ball? There is nothing wrong with *incommunicable intuitions*, but to relate them to obviousness is of a more shaky nature. – Mariano Suárez-Álvarez Sep 04 '11 at 20:08
  • @Asaf: I would bet Euler would disagree with your characterization of his understanding. Euler published according to the standards of his day and that has little to do with his actual understanding of things. I'm pretty sure he would have had no trouble matching modern publication standards but that wasn't the fashion, so he didn't bother. Intuition is far more than an emotion -- intuition is the starting point of real understanding, and quite often proofs. – Ryan Budney Sep 04 '11 at 20:12
  • @Mariano: I don't believe that I have ever heard about it being generalized to higher dimension, until now that is. I admit it is harder to have intuition when it comes to higher dimension; however thinking about it - the idea seems as a valid idea. I honestly can't tell about you. However as someone who never studied deep into geometry, of any kind (mostly because my geo. thinking is not my strong suit), I can say that it seems like a reasonable theorem. If you and Ryan think differently, I can't really help you with that. – Asaf Karagila Sep 04 '11 at 21:53
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    @Asaf: the interesting point is that it is a *false* statement (that the interior of a sphere is a ball) – Mariano Suárez-Álvarez Sep 04 '11 at 22:06
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    @Mariano: If all open sets were balls, life were a lot simpler (plus topology was not closed under unions :-)). The intuition is coming from simpler forms. If you are actually doing mathematics *without any intuition* then I am honestly amazed by the levels of formal thinking that you are able to carry out. Personally, I have to be guided by some informal intuition. – Asaf Karagila Sep 04 '11 at 22:09
  • @Mariano, marking this as the accepted answer. but still there are a few things that i might need to clarify. hopefully, i will contact you again once i have worked a little more on your suggested idea. – Akash Kumar Sep 05 '11 at 00:45
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    @Asaf: for me the remarkable thing is there's nothing about your intuition that you seem to be willing to articulate. Usually when someone says something is intuitive they can follow that up with some sort of idea as to why something should be true, even if half-baked. I *do* believe it is intuitive, but the intuition is based on an actual idea that leads to a proof. – Ryan Budney Sep 05 '11 at 01:41
  • @Ryan: The truly remarkable thing is that in the now-closed "surprising results" as all the links appearing in the comments (which include MO/MSE favorite/surprising generalizations lists and so on), which are threads both yourself and Mariano took part in. In all those answers and comments appearing in all those pages, not a single time the word "Jordan" appears. And you know why? Because you don't expect it not to be true without *much* prior knowledge. However, I do agree that studying a lot destroys a prior intuition in favor of new one, which may explain why you see this as non-intuitive. – Asaf Karagila Sep 05 '11 at 23:04
  • @Ryan: When the theorem was announced no one fell of his chair, no one held his hand and found his mathematical view of the world is shattered, no one wrote "I see it, but I don't believe it!". And if anything like that happened, well then.. this is remarkable. – Asaf Karagila Sep 05 '11 at 23:16
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    I wouldn't say studying destroys any intuition -- it can destroy preconceptions. But to me a preconception/bias is quite different from an intuition. In my mind there's a sliding scale from intuition to "hunch" to preconception/bias. – Ryan Budney Sep 05 '11 at 23:58

I think perhaps the biggest thing that blinds one's intuition is that when one imagines embedding a circle in the plane, it's very easy to "lose the plot" and instead imagine embedding a disc in the plane, with the circle on the boundary. So not only do you see immediately the inside and outside, but you also see the Schoenflies theorem -- that a circle in the plane bounds a disc.

A game that might help you check your intuition would be to get someone else to draw an extremely complicated simple closed curve in the plane, and then see if you can determine whether or not it has an "inside" and "outside" quickly. In particular, see if you can quickly judge whether or not parts of the diagram are in the "inside" and "outside", and then check more carefully to see if you were correct.

This game can be made more complicated by there being more than one disjoint simple closed curve in the diagram. What happens then? And so on. I suspect that until you find the germ of the proof, this puzzle has much the same shape as the kind of "maze puzzles" you see in comic sections of newspapers.

Ryan Budney
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The Jordan Curve theorem is actually pretty easy to prove if you assume the curve is smooth or piecewise linear. The difficulty arises when you try to handle the general case. This includes nowhere-differentiable curves like the boundary of the Koch snowflake, and even wilder curves which can't even be drawn by hand, like Mariano says. It's kind of a magic fact about the plane that simple closed curves behave nicely enough for the JCT or Schoenflies theorem to hold. Actually JCT holds for any sphere $S^n$ embedded in $\mathbb R^{n+1}$ by Alexander duality, but the Schoenflies theorem does not (when $n>2$). Alexander's horned sphere is an example of $S^2$ embedded in $\mathbb R^3$ whose complement is not simply connected. Another phenomenon that "magically" doesn't happen in the plane is the phenonemon of wildness. There are embedded compact arcs in $\mathbb R^3$, such as the Fox-Artin wild arc, which have non-simply connected complements. So you can never straighten these arcs out. However in the plane any arc (or simple closed curve) can be straightened out by an ambient isotopy.

Edit: I am including a sketch of the proof in the case of a polygonal curve, following Stillwell. The smooth case can be proven by a similar argument.

Let $p$ be a polygonal curve and let $\mathcal N$ be a tubular neighborhood. Then $\mathcal N\setminus p$ has two components. (Otherwise it would be a Möbius band, which would give a non-orientation preserving path in the plane, a contradiction.) So $\mathbb R^2\setminus p$ has at most $2$ components since any point in $\mathbb R^2\setminus p$ can be joined to $\mathcal N$ by a path.

Now we show that there are exactly two components. Pick a direction which is not parallel to any segment of $p$, and consider the family of lines $\{\mathcal \ell\}$ in the plane parallel to that direction. Define the set $\mathcal O$ to be the set of points which lie on the unbounded component of any $\ell\setminus p$, or which are an even number of crossings from the unbounded component. Define $\mathcal I$ to be the set of points which are an odd number of crossings away from the unbounded component. The crucial lemma is that each of these sets is open. (Note that touching a vertex does not count as a crossing.) This can be seen by examining the ways in which the line $\ell$ can meet the polygon. The only somewhat subtle case is when $\mathcal \ell$ meets a vertex and both adjacent segments lie to one side of $\ell$.

Now we conclude that $\mathcal O$ and $\mathcal I$ are a separation of $\mathbb R^2\setminus p$, so the complement has at least two components.

Cheerful Parsnip
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    Do you have a reference for a proof of the smooth JCT? I've been trying to remember where I saw it last time, with no luck. – Mariano Suárez-Álvarez Sep 05 '11 at 03:53
  • I think John Stillwell has a proof in his book "Classical Topology and Combinatorial Group Theory." Maybe that's just the PL case. – Cheerful Parsnip Sep 05 '11 at 12:46
  • Yup. He does the PL case. – Mariano Suárez-Álvarez Sep 05 '11 at 15:14
  • @Mariano: Once you can prove that the curve has a regular neighborhood which is a normal bundle, then Stillwell's proof should still work. I suppose it's debatable about whether proving the existence of such a neighborhood is significantly simpler than proving the full JCT in all its glory. – Cheerful Parsnip Sep 05 '11 at 16:23
  • I'll check when I get hold of the book, but I think Stillwell works by induction on the number of sides in the polygon, showing one can always shorten it by one (and thus proving Schoenflies theorem along the way) I'd have to think a bit how a normal neighborhood helps us do something similar in the smooth case. – Mariano Suárez-Álvarez Sep 05 '11 at 17:00
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    The smooth JCT is an exercise on Page 3 of Hatcher's "Notes on Basic 3-Manifold Topology". It's a nice, easy exercize in basic Morse Theory. It is proved the same way that he proves the Alexander Theorem, but it's much easier...The only fiddly point is "rounding the corners" when you slice the curve up into smaller curves. – Daniel Moskovich Nov 15 '11 at 15:07
  • @DanielMoskovich thanks for the reference! – Cheerful Parsnip Nov 15 '11 at 15:11
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    "The Jordan curve theorem is actually pretty easy to prove if you assume the curve is smooth or piecewise linear." I guess you don't know what you are talking about. Otherwise, I urge you to write the proof for a Jordan polygon here. If it's pretty easy, you can do it without a hassle. – Makoto Kato Aug 01 '16 at 22:50
  • @MakotoKato: "Easy" is a relative term I guess. Perhaps I should have said "much easier" than the general case. Stillwell gives a proof for the PL case and Daniel Moskovich gives a reference to Hatcher for the smooth case. Both proofs are much simpler than the general argument. On the other hand, I just recently taught the proof given by Munkres for the general case, and in fact it is quite nice. However you need to know fundamental groups and covering spaces. – Cheerful Parsnip Aug 02 '16 at 03:15
  • @GrumpyParsnip It would be nice if you show a rough sketch of the proof for a simple polygon. – Makoto Kato Aug 02 '16 at 03:26
  • @MakotoKato, I added a sketch of the proof found in Stillwell's book. Also check out p.26 of http://www.maths.ed.ac.uk/~aar/papers/stillwell1x.pdf – Cheerful Parsnip Aug 02 '16 at 06:51
  • @GrumpyParsnip Great. Thanks. – Makoto Kato Aug 02 '16 at 06:55

This is an old question, but here are some visual examples that help to suggest that indeed the Jordan curve theorem is actually not so intuitively obvious.

Both are not rectifiable. We have relatively good intuition about rectifiable curves but not non-rectifiable ones. These examples may seem simple, but imagine sprouting lots of similar smaller pieces along the way like a fractal. For the blue one, imagine the 'arms' intertwining more and more the deeper you go in. It would still be continuous but would be less obvious why there might not just be a way to divide the plane into three or more parts!

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    How do you say that either curve is a continuous loop?? – curiousdannii Aug 18 '17 at 12:48
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    @curiousdannii: They are both continuous (despite being not rectifiable); each has a continuous parametrization. If you don't get it, try the first one first: there are countably many 'rounds' so you can fit each into a rational interval leading up to zero from both sides. – user21820 Aug 18 '17 at 20:07
  • I should have asked how they can be closed, for isn't that required for the JCT? – curiousdannii Aug 18 '17 at 23:42
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    @curiousdannii: Yes that is a requirement for the JCT. The diagrams are merely to illustrate how a portion of the curve might look like. You can join up the ends to get a closed continuous curve, and it is not really obvious that it partitions the plane into a distinct 'inside' and 'outside'. Sorry I didn't know you were asking about the "loop" part. – user21820 Aug 19 '17 at 09:54