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Does there exist a piecewise smooth curve in $\mathbb{R}^3$ such that every plane intersects the curve at finitely many points and the number of intersection points can be arbitrary large?

If the number of intersection points for each plane is bounded then there is an example: $\gamma(t)=(t,t^3,t^5)$ intersects every plane at most five points.

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What does that look like? I tried to get Wolfram|Alpha to plot it, but all I could get it to do is plot the three curves $y=x, y=x^3, y=x^5$. – MJD Jul 25 '12 at 16:11
I also don't know what gamma looks like.I just show that the equation $at+bt^3+ct^5+d=0$ always has at least one solution and at most five soution. – Ben Jul 25 '12 at 16:16
@Mark, not that interesting-looking at first glance... – J. M. Jul 25 '12 at 16:32
Perhaps something like $t\mapsto (t+\frac{\sin(t^5)}{1+t^2},t^3,t^5)$? – Henning Makholm Jul 25 '12 at 16:32
If the number of intersection points can be arbitrarily large, then your space curve certainly cannot be algebraic (i.e. the intersection of two algebraic surfaces). I think there might be a trigonometric function component, but those things tend to intersect at infinitely many points... – J. M. Jul 25 '12 at 16:34
up vote 3 down vote accepted

Try $(x,y,z) = (t + \sin(t^2), t^2, t^3)$. For any $a,b,c$ (not all $0$) and $d$, $|a (t+\sin(t^2)) + b t^2 + c t^3 - d| \to \infty$ as $t \to \pm \infty$ so the intersections are in a finite interval. And since $a (t+\sin(t^2)) + b t^2 + c t^3 - d$ is analytic and not constant, it has finitely many zeros in a compact set. So any plane has only finitely many intersections with the curve.

The curve intersects the plane $x = \sqrt{2 m \pi}$ (where $m$ is a positive integer) when $t + \sin(t^2) = \sqrt{2 m \pi}$. For $t = \sqrt{2m \pi}+s$ that says $$s + \sin((\sqrt{2m\pi}+s)^2) = s + \sin(2 \sqrt{2m\pi} s + s^2) = 0$$ In the interval $-1/2 < s < 1/2$, $2 \sqrt{2m\pi}s +s^2$ runs from $-\sqrt{2m\pi} + 1/4$ to $+\sqrt{2m\pi} + 1/4$, and thus passes through approximately $\sqrt{2m/\pi}$ odd multiples of $\pi/2$, at which the sine is alternately $\pm 1$, and thus $ s + \sin(2 \sqrt{2m\pi} s + s^2)$ has approximately $\sqrt{2m/\pi}$ sign changes. Thus the number of intersection points is unbounded.

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Why does "any plane has only finitely many intersections with the curve" follow from "$f(t)$ has finitely many zeroes in a compact set"? – MJD Jul 25 '12 at 20:39
All the zeros are in a compact set, and there are only finitely many in that compact set. – Robert Israel Jul 26 '12 at 0:12
Oh, so it's "…has finitely many zeroes, in a compact set", not "…has finitely many zeroes in any given compact set"? – MJD Jul 26 '12 at 0:14
In any given compact set, it has only finitely many zeros. – Robert Israel Jul 26 '12 at 3:00

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