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I posted a question on math.stackexchange.com but it seems this question might be open https://math.stackexchange.com/questions/109752/line-segments-intersecting-jordan-curve

Namely,

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?

In the link Leonid Kovalev gave an example that might solve the problem but I have no idea is that an example of such curve. Can anyone verify if the Julia set he gave in the link solves the problem?

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    $\begingroup$ Take a circle. Then modify it by deforming the curve into a sequence of interlocking hooks, a bit like a zip. Then repeat this procedure with smaller hooks, and repeat so that the curve approaches a fractal. That will give you an example. $\endgroup$
    – George Lowther
    Commented Jun 19, 2012 at 18:56
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    $\begingroup$ See my icon for an example. $\endgroup$
    – Ian Agol
    Commented Jun 19, 2012 at 19:34
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    $\begingroup$ Given any modulus of continuity, you can construct the jordan curve in such a way that every curve with the given modulus of continuity joininng the inside of the region to the outside intersects the Jordan curve infinitely often. So you can replace "line segment" by Lipschitz continuous curve or Holder continuous curve if you like. $\endgroup$
    – George Lowther
    Commented Jun 19, 2012 at 19:34

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Most* of Jordan curves have this property. Moreover for most of curves you can not see a point on curve from one side if you can see it from the other side.

To construct such example, start with a smooth closed curve $\gamma_0$, note that you can wave it, so it will remain smooth and any point which is visible from one side on distance 1 is not visible from the other side on distance 1.

alt text

Given $\varepsilon_1>0$, you may assume in addition that the new curve $\gamma_1$ is $\varepsilon_1$-close to $\gamma_0$; i.e. $$|\gamma_1(t)-\gamma_0(t)|<\varepsilon_1$$ for any $t$.

Repeat this procedure for distances $\tfrac12$, $\tfrac13$ and so on. At each step, choose $\varepsilon_n$ very small, depending on $\gamma_{n-1}$, then the limit $\gamma_\infty$ is a Jordan curve you want.

$*$ "Most" means a G-delta dense set of Jordan curves.

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I might be wrong, but the Newton fractal: http://en.wikipedia.org/wiki/File:Julia_set_for_the_rational_function.png might be a good candidate.

See also http://en.wikipedia.org/wiki/Lakes_of_Wada

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  • $\begingroup$ the lakes are not bonded by a Jordan curves... $\endgroup$
    – Anton Petrunin
    Commented Jun 20, 2012 at 15:39
  • $\begingroup$ Yes, that is correct, but one might be able to modify it a bit maybe? Taking one open component, and its boundary as the curve... $\endgroup$
    – Per Alexandersson
    Commented Jun 21, 2012 at 7:52
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    $\begingroup$ No, that is the point, the lakes are not bounded by Jordan curves; otherwise any lake would contradict Jordan curve theorem. $\endgroup$
    – Anton Petrunin
    Commented Jun 30, 2012 at 18:14

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