Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Can a smooth closed real plane curve intersect itself at infinitely many points? It seems intuitively obvious that the answer should be no, yet I have no idea how to prove this or construct a counter-example. Here by smooth I mean $C^1$. If the answer is no, to which $C^k$ do we have to move for this geometric condition to be satisfied?

Edit: Here is an attempt to formalize the above: Let $C$ be a closed curve and $P$ a point at its image. We say that $C$ intersects itself at P, if for all parametrizations $f: [a,b] \to C$ (which are of the same $C^k$ class as C), the equation $f(x)=P$ has at least two solutions in $[a,b]$. I think this would work for what I had in mind posing this question.

By the way, I have no idea if this is the same with the transversal intersection definition proposed below.

share|improve this question
3  
Isn't $\gamma : \mathbb{R} \to \mathbb{R}^2$, defined by $\gamma (t) = (\cos(t), \sin(t))$, a closed curve in $\mathbb{R}^2$ that intersects itself at infinitely many points? –  Rod Carvalho Sep 11 '12 at 14:23
    
Huh... yes technically, but I mean a "geometric" self-intersection; e.g the curve that looks like an 8 has 1 self-intersection. Perhaps you could say the normal and tangent vector cannot coincide on open intervals, but I am not sure if this gives precisely what I mean. But I am sure you can visualize what I mean? –  Bernard Sep 11 '12 at 14:30
    
Let me suggest another variant of the question: "Can a smooth closed real plane curve intersect itself at infinitely many points with all self-intersections being transversal?" Seems the answer is "no" now. –  scholar Sep 11 '12 at 14:45
1  
I guess you only want to count "transverse intersections". en.wikipedia.org/wiki/Transversality_%28mathematics%29 –  j.c. Sep 11 '12 at 14:46
    
If you keep the intersections apart from each other you can prevent this because the curve is a continuous image of a compact set, therefore compact. The spaces between the intersections form an infinite (almost) cover with no finite subcover. –  Ross Millikan Sep 11 '12 at 14:56

2 Answers 2

up vote 2 down vote accepted

Here is a less trivial example. The function

$$f(x)=\begin{cases} 0&\quad \text{if} \quad x=0\\ x^p \sin(1/x) &\text{otherwise} \end{cases}$$

is as smooth as you want (making $p$ large) but intersects the zero line infinitly often for $x\in[0,1]$. From this function you can easily make a closed loop intersecting itself infinitely often this way.

share|improve this answer
    
OK... how about analytic functions? Does it work there? –  Bernard Sep 11 '12 at 14:43
    
@Bernard Yes, it does. An infinite set of self intersections would have a limit point, forcing (by the identity theorem) the curve to be periodic, like the parametrization of circle mentioned above. –  user31373 Sep 12 '12 at 5:10

How about $y=x^2 \sin \frac 1x$ and $y=0$ on $x \in [0,1]$ plus a smooth turnaround at each end?

share|improve this answer
    
hmm, you beat me to it:) –  Thomas Rot Sep 11 '12 at 14:36
    
And if you want $C^{\infty}$, just use $e^{-1/x^2} sin (1/x)$ instead. –  David Speyer Sep 11 '12 at 15:08
    
@DavidSpeyer: I don't think changing the continuity level makes it easier or harder once you get past $C^1$ as you say. –  Ross Millikan Sep 11 '12 at 15:13

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.