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Let us say we have a polygon $P$ in $\mathbb{R}^2$, with edges in the set $E$ (the boundary), and vertices in the set $V$. Let us say we have a point $Q$ such that $Q$ lies on one of the edges in $E$.

Let the direction of a vector not parallel to any of the edges in $E$ be denoted as $\alpha$. Consider also $\beta = \alpha+ \pi$. Note that a vector with angle $\beta$ is also not parallel to any of the edges in $E$. Given some direction angle $\gamma$, let the ray emanating from $q$ be denoted as $R_{\gamma}$

Let us consider then that direction $\alpha$ points "out of the polygon", while direction $\beta$ points "into the polygon". Then, we see that the ray $R_{\alpha}$ intersects with the $E$ once (at $q$), while the ray $R_{\beta}$ intersects with $E$ twice (once at $q$, and once more when the ray exits the polygon).

So, the discrete Jordan curve theorem cannot talk about points on the boundary of $P$?

bzm3r
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1 Answers1

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Jordan curve theorem states that if $C\subset R^2$ is homeomorphic to $S^1$ then $R^2\setminus C$ consists of two components, one bounded, $V$, and one unbounded, $U$. Note that there is nothing here about points of $C$ since $C$ is removed. Jordan's theorem is supplemented by Schoenflies theorem which states that $V\cup C$ is homeomorphic to the closed disk.

Moishe Kohan
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  • I am afraid I can't make much sense of your answer (but that is my fault). What do you mean by $S^{1}$? – bzm3r Jun 17 '14 at 16:13
  • $S^1$ is the standard notation for the unit circle. (Similarly, $S^n$ stands for the unit sphere.) In general, my suggestion is to read the actual proof of the Jourdan separation theorem (at least in the polygonal case) instead of trying to invent your own. – Moishe Kohan Jun 17 '14 at 16:18