User:Anghel/Sandbox

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 components.

Both components are open in $\R^2$ and have $\partial P$ as their toplogical boundary.

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

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

Lemma 1
$\R^2 \setminus \partial P$ is a union of at most two disjoint path-connected sets.

Proof
The polygon $P$ has $n$ sides, where $n \in \N$.

Denote the vertices of $P$ as $A_1, \ldots, A_n$ and its sides as $S_1, \ldots, S_n$, such that each vertex $A_i$ has adjacent sides $S_{i-1}$ and $S_i$.

We use the conventions that $S_0 = S_n$, and $A_{n+1} = A_1$.

Let $\displaystyle \delta_i = d \left({A_i, \bigcup_{ j = 1, \ldots, n, \ j \ne i-1, j \ne i } S_j }\right)$ be the Euclidean distance between a vertex $A_i$ and the sides not adjacent to $A_i$.

From Distance between Closed Sets in Euclidean Space, it follows that $\delta_i > 0$

Put $\displaystyle \delta = \min_{i=1, \ldots, n} \delta_i$.

From Boundary of Polygon is Jordan Curve, it follows that $\partial P$ is equal to the image of a Jordan curve $\gamma: \left[{0\,.\,.\,1}\right] \to \R^2$ that is a concatenation of $n$ paths $\gamma_1, \ldots, \gamma_n$.

Each $\gamma_i$ is a line segment that joins its initial point $A_i$ and its final point $A_{i+1}$, so the image of $\gamma_i$ is equal to the side $S_i$.

Let $\mathbf v_i = \dfrac{A_{i+1} - A_i}{d \left({A_{i+1}, A_i}\right) }$ be the direction vector of $\gamma_i$ with norm $\left\Vert{\mathbf v_i}\right\Vert = 1$.

Let $\mathbf w_i$ be the vector $v_i$ rotated $\dfrac \pi 2$ radians counterclockwise, so $\left\Vert{\mathbf w_i}\right\Vert = 1$.

For any $\epsilon \in \left({0\,.\,.\,\dfrac \delta 2}\right)$, we intend to construct two Jordan curves $\sigma, \overline \sigma$ such that $\operatorname{Im} \left({\sigma}\right) \cup \operatorname{Im} \left({\overline \sigma}\right) = \left\{ {q \in \R^2: d \left({q, \partial P}\right) = \epsilon }\right\}$.

For $i \in \left\{ {1, \ldots, n}\right\}$, initially define the $\sigma_i$ as the line segment that joins its initial point $A_i + \epsilon \mathbf w_i$ with its final point $A_{i+1} + \epsilon \mathbf w_i$.

If $\sigma_i$ and $\sigma_{i+1}$ intersect at some point $p_{i+1} \in R^2$, re-define the two line segments so the final point of $\sigma_i$ becomes $p_{i+1}$, and the initial point of $\sigma_{i+1}$ becomes $p_{i+1}$.

Then define a path $\rho_i$ as the constant function $\rho_i \left({t}\right) = p_{i+1}$.

Otherwise, define a path $\rho_i$ with initial point $A_{i+1} + \epsilon \mathbf w_i$ and final point $A_{i+1} + \epsilon \mathbf w_{i+1}$, such that the image of $\rho_i$ is part of the circumference of the circle with center $A_{i+1}$ and radius $\epsilon$.

Define the path $\sigma: \left[{0\,.\,.\,1}\right] \to \R^2$ as the concatenation:


 * $\sigma = \sigma_1 * \rho_1 * \sigma_2 * \rho_2 * \ldots \sigma_n * \rho_n$

Then, $\sigma$ is a closed path, as $\sigma_1$ has initial point $A_1$ equal to the final point of $\rho_n$.

Each of the paths $\sigma_i$ and $\rho_i$ are injective.

For all $i, j \in \left\{ {1, \ldots, n}\right\}$, $\sigma_i$ only intersects $\rho_{i-1}$ and $\rho_i$ in their endpoints.

Also, $\sigma_i$ can only possibly intersect $\sigma_{i-1}$ in their endpoints, in which case the path $\rho_{i-1}$ is constant.

Similarly, $\sigma_i$ can only possibly intersect $\sigma_{i+1}$ in their endpoints, in which case the path $\rho_{i+1}$ is constant.

For $\left\vert{i - j}\right\vert > 1$, $\sigma_i$ and $\sigma_j$ cannot intersect.