I have thought this problem a week without success.

Is there a set $A\subset \mathbb{R}^2$ such that

  • The boundary of $A$, $\partial A$, is a Jordan curve and
  • For any $B\in \operatorname{int} A\ne\emptyset $, $C\in \operatorname{ext} A\ne\emptyset$ , the line segment $BC$ intersects $\partial A$ infinitely many times?

Any ideas?

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    $\begingroup$ Fascinating question. If there is such an example, the intersection of the Jordan curve with any line cannot contain an isolated point. For otherwise you could take a point to either side of the isolated point and get a line segment joining inside and outside which hits the curve once. Also the curve is closed, so the intersection set with any line must be a closed subset of the line with no isolated points. This means it is a perfect set, and a theorem says that it is therefore uncountable. $\endgroup$ – Cheerful Parsnip Feb 16 '12 at 2:40
  • $\begingroup$ So an example must hit every line joining interior and exterior uncountably many times. $\endgroup$ – Cheerful Parsnip Feb 16 '12 at 2:41
  • $\begingroup$ I have very little intuition about this; can a line segment intersect the boundary of the Koch snowflake finitely many times? $\endgroup$ – Miha Habič Feb 16 '12 at 9:50
  • $\begingroup$ @MihaHabič: yes, it can hit it in one point at a "corner." $\endgroup$ – Cheerful Parsnip Feb 16 '12 at 12:13
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    $\begingroup$ $A=\partial A=S^1$ works fine, but you'd probably prefer $\operatorname{int} A$ to be nonempty ;) $\endgroup$ – savick01 Feb 18 '12 at 23:29

This is not an answer, only some ideas.

Let $A$ be a Jordan domain, and let $f\colon \mathbb D\to A$ be a conformal map of the unit disk onto $A$. By Caratheodory's theorem $f$ extends to a homeomorphism of closures. One says that $f$ is twisting at a point $\zeta\in\partial\mathbb D$ if for every curve $\Gamma\subset \mathbb D$ ending at $\zeta$ the following holds: $$\liminf_{z\to\zeta}\ \arg(f(z)-f(\zeta))=-\infty, \quad \limsup_{z\to\zeta}\ \arg(f(z)-f(\zeta))=+\infty,\quad (z\in\Gamma) $$

(This definition is from the book Boundary behaviour of conformal maps by Pommerenke. Some sources have different wording. The definition is unchanged if one replaces $\Gamma$ by a radial segment.)

If $f$ is twisting at $\zeta$, there is no line segment that crosses $\partial A$ only at $f(\zeta)$. Indeed, the preimage of such a line segment would be a curve along which $\arg(f(z)-f(\zeta))$ is constant.

Pommerenke's book presents several results on twisting points and gives pointers to literature. The message is that $f$ can have a lot of twisting points. Of course, it is impossible for $f$ to be twisting at every point of $\partial\mathbb D$. (Consider any disk contained in $A$ whose boundary touches $\partial A$.) But what we want is for every point of $\partial A$ to be twisting either on the inside or on the outside (i.e., for the conformal map onto interior or onto exterior).

My profound lack of knowledge of complex dynamics suggests that the Julia set of the quadratic polynomial $p(z)=z^2+\lambda z$ could have this property when $|\lambda|<1$ and $\mathrm{Im}\,\lambda\ne 0$. Indeed, in this case the polynomial has two Fatou components, and the boundary between them is a Jordan curve, indeed a quasicircle. The curve appears to be twisting as expected (see this applet, which takes polynomials in the form $z^2+c$. Here $c=\lambda/2-\lambda^2/4$ with $|\lambda|<1$, so we are in the main cardioid of the Mandelbrot set). Maybe someone who understands complex dynamics can tell if this Julia set is indeed an example.


I am not sure, but here's an idea. Take the graph of $x sin(1/x)$ and join the two zeroes around (+/-)0.15 by a big enough curve. Take the points to be $(0.12, 0)$ and $(-100, 0)$.

EDIT: Misunderstood, that it is required for only a single pair of points. OP wants this to happen for every pair of points.

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    $\begingroup$ I also thought the solution might be some kind of fractal. $\endgroup$ – Jaakko Seppälä Feb 15 '12 at 21:21
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    $\begingroup$ I believe the OP wants a curve which works for every pair of points. $\endgroup$ – Cheerful Parsnip Feb 16 '12 at 2:29
  • $\begingroup$ That's right Jim. I'm non-native in English and I thought that for any means for all. $\endgroup$ – Jaakko Seppälä Feb 16 '12 at 9:40
  • $\begingroup$ Actually, rereading the question I can see this. I misunderstood.. My mistake! Sorry. $\endgroup$ – aelguindy Feb 16 '12 at 9:45
  • $\begingroup$ @JaakkoSeppälä: "for any" does mean "for every" in most situations including your question (so it is not wrongly phrased). A situation where "for any" would be ambiguous is "if for any ...". $\endgroup$ – user21820 Oct 26 '15 at 8:06

Such a set does exist. See https://mathoverflow.net/questions/100025/how-many-times-line-segments-can-intersect-a-jordan-curve

  • $\begingroup$ Nice construction by Anton. You should accept this as an answer instead of my half-baked remarks. $\endgroup$ – user31373 Jun 19 '12 at 20:42

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