Jordan Polygon Theorem

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

Denote the boundary of $P$ as $\partial P$.

Then, $\R^2 \setminus \partial P$ is a union of two connected components.

Both components are open in $\R^2$.

One component is bounded, and is called the interior of $P$.

The other component is unbounded, and is called the exterior of $P$.

Lemma
We show that $\R^2 \setminus \partial P$ is not path-connected.

Find any $q_1 \in R^2 \setminus \partial P$ and $\theta \in \R$ such that the ray $\mathcal L_\theta = \left\{ {q_1 + s g \left({\theta}\right): s \in \R_{\ge 0} }\right\}$ has exactly one crossing of $\partial P$.

Find any $q_2 \in \mathcal L_\theta$ that lies on the ray after the crossing, so the ray $\left\{ {q_2 + s g \left({\theta}\right): s \in \R_{\ge 0} }\right\}$ does not intersect $\partial P$.

Then $\operatorname{par} \left({q_1}\right) = 1 \ne 0 = \operatorname{par} \left({q_2}\right)$.

From Jordan Polygon Parity Lemma, it follows that $q_1$ and $q_2$ cannot be connected by a path.

As $R^2 \setminus \partial P$ is not path-connected, it follows from the Lemma that $R^2 \setminus \partial P$ is a union of exactly two disjoint path-connected sets, which we denote as $U_1$ and $U_2$.

Let $q \in \R^2 \setminus \partial P$, and let $d \left({q, \partial P}\right)$ be the Euclidean distance between $q$ and $\partial P$.

From Distance between Closed Sets in Euclidean Space, it follows that $d \left({q, \partial P}\right) > 0$.

When we put $\epsilon = d \left({q, \partial P}\right) / 2$, we have $B_\epsilon \left({q}\right) \subseteq \R^2 \setminus \partial P$.

As Open Ball is Convex Set, it follows that $B_\epsilon \left({q}\right)$ is path-connected, so $B_\epsilon \left({q}\right)$ is a subset of either $U_1$ or $U_2$.

Then, both $U_1$ and $U_2$ are open.

From Path-Connected Space is Connected, it follows that $U_1$ and $U_2$ are connected.

Then, $\R^2 \setminus \partial P$ is a union of two components.

From Boundary of Polygon is Jordan Curve, it follows that $\partial P$ is equal to the image of a Jordan curve $\gamma: \left[{0 \,.\,.\, 1}\right] \to \R^2$.

From Continuous Image of Compact Space is Compact/Corollary 2, it follows that $\partial P$ is bounded.

That is, there exist $a \in \R^2$ and $R \in \R_{>0}$ such that $\partial P \subseteq B_R \left({a}\right)$.

If $x_1, x_2 \in \R^2 \setminus B_R \left({a}\right)$, $x_1$ to $x_2$ can be joined by a path in $\R^2 \setminus B_R \left({a}\right)$ following:


 * the circumference of the two circles with center $a$ and radii $d \left({a, x_1}\right)$ and $d \left({a, x_2}\right)$.


 * a line segment joining the two circumferences

Then $\R^2 \setminus B_R \left({a}\right)$ is path-connected, so $\R^2 \setminus B_R \left({a}\right)$ is a subset of one of the components of $\R^2 \setminus \partial P$, say $U_1$.

As $\R^2 \setminus B_R \left({a}\right) \subseteq U_1$, it follows that $U_1$ is unbounded, so $U_1$ is the exterior of $\gamma$.

Then $U_2 \subseteq B_R \left({a}\right)$, so $U_2$ is bounded, which implies that $U_2$ is the interior of $\gamma$.