# Why is the Jordan Curve Theorem not "obvious"?

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.

• @Akash: What are the "obvious reasons" for its truth? Sep 4, 2011 at 17:09
• 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! Sep 4, 2011 at 17:10
• 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 $\mathbb R^4$? Sep 4, 2011 at 17:11
• 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 4, 2011 at 18:34
• That lots of "obvious" things are false may be seen by looking at the book Counterexamples in Analysis. Sep 4, 2011 at 21:21

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)

• @Asaf: what do you mean by intuitively obvious? I think it helps to be specific about what intuitive mechanisms you're using. Sep 4, 2011 at 17:10
• @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. Sep 4, 2011 at 19:29
• @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. Sep 4, 2011 at 19:57
• @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. Sep 4, 2011 at 20:08
• @Asaf: the interesting point is that it is a false statement (that the interior of a sphere is a ball) Sep 4, 2011 at 22:06

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.

• There's an exercise like this at the beginning of Thurston's "Three dimensional topology and geometry." Sep 4, 2011 at 17:29

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.

• 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. Sep 5, 2011 at 3:53
• I think John Stillwell has a proof in his book "Classical Topology and Combinatorial Group Theory." Maybe that's just the PL case. Sep 5, 2011 at 12:46
• Yup. He does the PL case. Sep 5, 2011 at 15:14
• 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. Nov 15, 2011 at 15:07
• "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. Aug 1, 2016 at 22:50

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!

• How do you say that either curve is a continuous loop?? Aug 18, 2017 at 12:48
• @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. Aug 18, 2017 at 20:07
• I should have asked how they can be closed, for isn't that required for the JCT? Aug 18, 2017 at 23:42
• @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. Aug 19, 2017 at 9:54