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, which are both 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 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 straight line that goes through $A_i + \epsilon \mathbf w_i$ and $A_{i+1} + \epsilon \mathbf w_i$.

Then $\sigma_i$ is parallel to $S_i$ and will intersect $\sigma_{i+1}$ at some point $p_{i+1}$.

This follows as $\sigma_{i+1}$ is parallel to $S_{i+1}$, which is not parallel to $S_i$ by the definition of polygon.

If $d \left({p_{i+1}, A_{i+1} }\right) = \epsilon$, cut the lines so the final point of $\sigma_i$ becomes $p_{i+1}$, and the initial point of $\sigma_{i+1}$ becomes $p_{i+1}$.

Otherwise, if $d \left({p_{i+1}, A_{i+1} }\right) > \epsilon$, cut the lines so the final point of $\sigma_i$ becomes $A_{i+1} + \epsilon \mathbf w_i$, and the initial point of $\sigma_{i+1}$ becomes $A_{i+1} \epsilon \mathbf w_{i+1}$.

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