Jordan Polygon Interior and Exterior Criterion

Theorem
Let $P$ be a polygon embedded in $\R^2$.

Let $q \in \R^2 \setminus \partial P$, where $\partial P$ denotes the boundary of $P$.

Let $\mathbf v \in \R^2 \setminus \left\{ {\mathbf 0}\right\}$ be a non-zero vector, and let $\mathcal L = \left\{ {q + s \mathbf v: s \in \R_{\ge 0} }\right\}$ be a ray with start point $q$.

Let $N \left({q}\right) \in \N$ be the number of crossings between $\mathcal L$ and $\partial P$.

Then:


 * $(1): \quad$ $q \in \operatorname{Int} \left({P}\right)$, iff $N \left({q}\right) \equiv 1 \pmod 2$


 * $(2): \quad$ $q \in \operatorname{Ext} \left({P}\right)$, iff $N \left({q}\right) \equiv 0 \pmod 2$

Here, $\operatorname{Int} \left({P}\right)$ and $\operatorname{Ext} \left({P}\right)$ denote the interior and exterior of $\partial P$, when $\partial P$ is considered as a Jordan curve.

Proof
From Boundary of Polygon is Jordan Curve, it follows that $\partial P$ is equal to the image of a Jordan curve.

From the Jordan Polygon Theorem, it follows that $\operatorname{Int} \left({P}\right)$ and $\operatorname{Ext} \left({P}\right)$ are path-connected.

Then, Jordan Polygon Parity Lemma shows that $N \left({q}\right) = \operatorname{par} \left({q}\right)$, where $\operatorname{par} \left({q}\right)$ denotes the parity of $q$.

From Jordan Polygon Theorem, it follows that $\operatorname{Ext} \left({P}\right)$ is unbounded, while $\operatorname{Int} \left({P}\right)$ is bounded.

As $\partial P$ is the image of a Jordan curve, it follows from Continuous Image of Compact Space is Compact/Corollary 2 that $\partial P$ is also bounded.

Then, there exists $R \in \R_{>0}$ such that $\operatorname{Int} \left({P}\right) \cup \partial P \subseteq B_R \left({\mathbf 0}\right)$.

If $q_0 \in \R^2 \setminus B_R \left({\mathbf 0}\right)$, then $q_0 \in \operatorname{Ext} \left({P}\right)$.

Then, the ray $\left\{ {q_0 + s q_0: s \in \R_{\ge 0} }\right\} \subseteq \R^2 \setminus B_R \left({\mathbf 0}\right)$, so there are zero crossings between the ray and $\partial P$.

From Jordan Polygon Parity Lemma, it follows that $\operatorname{par} \left({q_0}\right) = 0$.

As $\operatorname{Ext} \left({P}\right)$ is path-connected, it follows from the lemma that for all $q \in \operatorname{Ext} \left({P}\right)$, we have $\operatorname{par} \left({q}\right) = 0$.

If $q_1 \in \R^2 \setminus \partial P$ with $\operatorname{par} \left({q}\right) = 1$, it follows that $q_1 \notin \operatorname{Ext} \left({P}\right)$.

As $\R^2 \setminus \partial P = \operatorname{Int} \left({P}\right) \cup \operatorname{Ext} \left({P}\right)$, it follows that $q_1 \in \operatorname{Int} \left({P}\right)$.

Again, Jordan Polygon Parity Lemma shows that for all $q \in \operatorname{Int} \left({P}\right)$, we have $\operatorname{par} \left({q}\right) = 1$.

So if instead $q_0 \in \R^2 \setminus \partial P$ with $\operatorname{par} \left({q}\right) = 0$, the only possibility is that $q_0 \in \operatorname{Ext} \left({P}\right)$.