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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.

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    $\begingroup$ 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? $\endgroup$ Commented Sep 11, 2012 at 14:23
  • $\begingroup$ 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? $\endgroup$
    – Bernard
    Commented Sep 11, 2012 at 14:30
  • $\begingroup$ 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. $\endgroup$
    – scholar
    Commented Sep 11, 2012 at 14:45
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    $\begingroup$ I guess you only want to count "transverse intersections". en.wikipedia.org/wiki/Transversality_%28mathematics%29 $\endgroup$
    – j.c.
    Commented Sep 11, 2012 at 14:46
  • $\begingroup$ 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. $\endgroup$ Commented Sep 11, 2012 at 14:56

2 Answers 2

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

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  • $\begingroup$ hmm, you beat me to it:) $\endgroup$
    – Thomas Rot
    Commented Sep 11, 2012 at 14:36
  • $\begingroup$ And if you want $C^{\infty}$, just use $e^{-1/x^2} sin (1/x)$ instead. $\endgroup$ Commented Sep 11, 2012 at 15:08
  • $\begingroup$ @DavidSpeyer: I don't think changing the continuity level makes it easier or harder once you get past $C^1$ as you say. $\endgroup$ Commented Sep 11, 2012 at 15:13
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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.

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  • $\begingroup$ OK... how about analytic functions? Does it work there? $\endgroup$
    – Bernard
    Commented Sep 11, 2012 at 14:43
  • $\begingroup$ @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. $\endgroup$
    – user31373
    Commented Sep 12, 2012 at 5:10

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