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My interest in this question comes from Cauchy's integral theorem. If you have a rectifiable Jordan curve, can any two points in its interior be connected by a path in the interior made up of a finite number of directed line segments?

Here's my heuristic for believing it's true: let $f(t), t \in [a, b]$ be a parametrization of the curve $J$. Place a mesh $(x_i)_{1 \le i \le n}$ on $[a, b]$ and construct a polygonal curve $P$ approximating $J$ by connecting $f(x_j)$ to $f(x_{j+1})$ with a directed line segment (connecting $f(x_n)$ to $f(x_1)$) and traversing the line segments in order. My guess is that as the mesh gets finer, eventually $P$ will lie in the interior of $J$ and will not cross $J$ anywhere. Since $P$ is a closed polygonal curve its interior can be triangulated, and since triangles are convex this means any two points in the interior of $P$ can be connected with a polygonal path. Let $A$ be the intersection of the exterior of $P$ with the interior of $J$. I also think that as the mesh gets fine enough any connected component of $A$ (of which there will be a finite number) will also be convex. Since the interior of $J$ is connected and covered by a finite number of convex sets the original conjecture will then follow. I can't demonstrate either of these heuristics formally though.

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Let $U\subseteq\mathbb{R}^n$ be any connected open set (e.g., the interior of a Jordan curve in $\mathbb{R}^2$). Then any two points in $U$ can be connected by a polygonal path $U$.

Indeed, if $x\sim y$ means "$x$ can be connected to $y$ by a polygonal path in $U$", then $\sim$ is an equivalence relation on $U$ (since you can reverse or concatenate polygonal paths). Moreover, if $x\in U$, then if $B$ is a ball around $x$ contained in $U$, $x\sim y$ for every $y\in B$. Thus each equivalence class of $\sim$ is open. The equivalence classes are thus a partition of $U$ into open sets. Since $U$ is connected, there can only be one equivalence class.

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  • $\begingroup$ Very smooth. I guess my approach was too explicitly geometric to be workable. Thanks! $\endgroup$ – Vik78 Sep 28 '16 at 3:28
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    $\begingroup$ Of course the polygonal path can be taken with each segment parallel to a coordinate axis, and this can even be useful in some applications. $\endgroup$ – bof Sep 28 '16 at 4:14
  • $\begingroup$ Is there a refinement of this to prove the existence of a non-self-intersecting path ? (an accessible reference would be sufficient). As is, transitivity would not immediately follow if trying to include non-self-intersection as a condition for equivalence. $\endgroup$ – Tom Collinge Aug 5 '18 at 11:39
  • $\begingroup$ @TomCollinge Yes. You can arrange that all the segments of the polygonal path have different slopes (if two have the same slope, replace one of them with two line segments that are very close to it but have slopes not used anywhere else), so it has only have finitely many self-intersections. Then you can just cut out all the redundant parts. (In other words, if the path hits the same point at time $s$ and time $t$, just skip the parts between $s$ and $t$. After making finitely many cuts like this, you get an injective path.) $\endgroup$ – Eric Wofsey Aug 5 '18 at 13:53
  • $\begingroup$ Awesome - thanks $\endgroup$ – Tom Collinge Aug 5 '18 at 14:13

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