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Let $S$ be a connected subset of $\mathbf{R}^3$. I would like to know if there is any condition on $S$ under which any straight line $p: [0, 1] \rightarrow \mathbf{R}^3$, where $p(0) \in S$ and $p(1) \notin S$, passes through the boundary of $S$ only finitely many times.

Here passing through may be defined as follows.

The curve $p$ is said to enter the boundary $\partial S$ from $A$ at time $t$ if $p(t) \in \partial S$ and there exists $\varepsilon > 0$ such that $p(t - \delta) \in A$ for all $0 < \delta < \varepsilon$.

The curve $p$ is said to leave the boundary $\partial S$ to $A$ at time $t$ if $p(t) \in \partial S$ and there exists $\varepsilon > 0$ such that $p(t + \delta) \in A$ for all $0 < \delta < \varepsilon$.

The curve $p$ passes through $\partial S$ if one of the following happens:

  • $p$ enters $\partial S$ from ext($S$) at time $t_0$ and leaves $\partial S$ to int($S$) at time $t_1$ where $p(t) \in \partial S, \forall t \in [t_0, t_1]$.

  • $p$ enters $\partial S$ from int($S$) at time $t_0$ and leaves $\partial S$ to ext($S$) at time $t_1$ where $p(t) \in \partial S, \forall t \in [t_0, t_1]$.

Note that $t_0$ may be equal to $t_1$.

I tried to think of it first in two dimensional. If I have a set $S$ whose boundary has some portion containing infinite wiggles (I think of if as something similar to the topologist's sine curve), there might be a case where a curve $p$, starting inside $S$ and ending outside, passes through the boundary infinitely many times.

$p$ passes through the boundary of $S$ infinitely many times

What conditions would prevent this situations from happening (in general)? I guess there may be some conditions on some kind of regularity of the boundary of the set $S$. Can anyone point me to some related keywords?

EDIT(1): Added the notion of passing through.

EDIT(2): Added an assumption on $p$.

EDIT(3): Restricted $p$ to be a straight line

EDIT(4): I think differentiability of the boundary curve (in two dimensional case, for example) might be related here. For example, from the figure above, I generated the boundary curve by attaching a segment of curve $y = x\sin(1/x)$, from $x = 0$ to some positive number, to some smooth curves. This boundary curve does not have continuous first derivative. I suspect that other boundaries with these kinds of infinite wiggles (still do know how to address it properly, though) would not have continuous derivative as well. Maybe if the boundary curve is $C^1-$ or $C^2-$continuous (and the path $p$ is a straight line), $p$ will then pass through the boundary of $S$ only finitely many times? Is there any related result? And can this idea be generalized to the three dimensional case as well?

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  • $\begingroup$ You might need to define 'pass through'. $\endgroup$ – Dan Rust Feb 15 '17 at 12:46
  • $\begingroup$ Thank you for your suggestion. I added the notion of passing through the boundary into the post. I wonder if there is a formal way to define so. What I added is only what I think of when I say the curve is passing through the boundary. $\endgroup$ – Petch Puttichai Feb 16 '17 at 2:12
  • $\begingroup$ I think you may need to restrict the family of curves you're considering, say, to have only finitely many segments --- have you tried constructing a piecewise-linear curve of finite length that has pathological behavior, a la the boundary of $S$ in your picture? $\endgroup$ – Neal Feb 16 '17 at 2:20
  • $\begingroup$ I see. I just realized that I forgot to mention that I assume that $p$ does not have that pathological behavior. So the only thing that could cause the problem would be the boundary of $S$. $\endgroup$ – Petch Puttichai Feb 16 '17 at 2:28
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    $\begingroup$ Ah okay. (Just to nitpick, part of the game is writing down what "pathological behavior, such as infinite wiggles" means. :) $\endgroup$ – Neal Feb 16 '17 at 2:32

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