Jordan Polygon Theorem/Lemma 1

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

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

$\R^2 \setminus \partial P$ is the 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 $A_1, \ldots, A_n$.

Denote the sides of $P$ 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 $\displaystyle \delta_i = d \left({S_i, \bigcup_{j \mathop = 1, \ldots, n: \ \left\vert{i - j}\right\vert > 1 } S_j }\right)$ be the Euclidean distance between a side $S_i$ and all sides not adjacent to $S_i$.

From Distance between Closed Sets in Euclidean Space:
 * $\delta_i > 0$

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

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 $A_{i+1}$.

Therefore 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 $A_{i+1} + \epsilon \mathbf w_i$.

Suppose $\sigma_i$ and $\sigma_{i+1}$ intersect at some point $p_{i+1} \in R^2$.

Then re-define the two line segments so that:
 * 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 $\rho_i$ is 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 its endpoint.

In that case the path $\rho_{i-1}$ is constant.

Similarly, $\sigma_i$ can only possibly intersect $\sigma_{i+1}$ in its endpoint.

In that case the path $\rho_{i+1}$ is constant.

For $\left\vert{i - j}\right\vert > 1$:
 * let $x_i \in \operatorname{Im} \left({\sigma_i}\right) \cup \operatorname{Im} \left({\rho_i}\right)$
 * let $x_j \in \operatorname{Im} \left({\sigma_j}\right) \cup \operatorname{Im} \left({\rho_j}\right)$.

Let $p_i, p_j \in \partial P$ be two points such that:
 * $d \left({x_i, p_i}\right) = d \left({x_j, p_j}\right) = \epsilon$

Suppose $x_i \in \operatorname{Im} \left({\sigma_i}\right)$.

Then we can put:
 * $p_i = x_i - \epsilon \mathbf w_i$

Suppose $x_i \in \operatorname{Im} \left({\rho_i}\right)$.

Then we can put $p_i = A_{i+1}$.

Then $x_i \ne x_j$, as:

It follows that $\sigma_i$ and $\rho_i$ do not intersect $\sigma_j$ and $\rho_j$.

It follows that $\sigma$ is a Jordan curve.

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

Proceed to define $\overline \rho_i$ and re-define $\overline \sigma_i$ similarly to what was done with $\sigma_i$ and $\rho_i$.

Finally, define $\overline \sigma$ as the concatenation:


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

It follows that $\overline \sigma$ is a Jordan curve as above.

We now show that $\sigma$ and $\overline \sigma$ do not intersect.

First, $\sigma_i$ and $\overline \sigma_i$ do not intersect, as they are both line segments parallel to $S_i$ with a distance of $2 \epsilon$.

Second, $\rho_i$ and $\overline \rho_i$ do not intersect, as one path, say $\rho_i$, is constant with image equal to the crossing point of $\sigma_{i-1}$ and $\sigma_i$.

The other path, say $\overline \rho_i$, is part of the circumference of the circle that joins the final point of $\overline \sigma_{i-1}$ with the initial point of $\overline \sigma_i$.

Third, $\sigma_i$ and $\overline \sigma_{i+1}$ do not intersect, since neither line segment crosses $\partial P$, which would be necessary if the two line segment should intersect.

Similarly, $\sigma_i$ and $\overline \sigma_{i-1}$ do not intersect.

For all other combinations of $i, j \in \left\{ {1, \ldots, n}\right\}$, $\sigma_i$ and $\rho_i$ do not intersect with $\overline \sigma_j$ and $\overline \rho_j$.

This follows as:
 * $d \left({ \sigma_i, S_i}\right) = d \left({ \rho_i, S_i}\right) = d \left({ \rho_i, S_{i-1} }\right) = \epsilon$

and as $S_i$, or $S_{i-1}$, are not adjacent sides to $S_j$, we have $d \left({S_i, S_j}\right) > 2 \epsilon$.

Now we can use the Triangle Inequality as above, to prove that the paths do not intersect.

Finally, let $q_1, q_2, q_3 \in R^2 \setminus \partial P$.

Let:
 * $\displaystyle \epsilon = \min \left({ d \left({q_1, \partial P}\right) / 2, d \left({q_2, \partial P}\right) / 2, d \left({q_3, \partial P}\right) / 2, \delta }\right)$

For all $i \in \left\{ {1, 2, 3}\right\}$, draw a line segment $\mathcal L_i$ joining $q_i$ with any point on the boundary $\partial P$.

As $d \left({q_i, \partial P}\right) > \epsilon$, it follows from the Intermediate Value Theorem that there is a point $x_i$ on $\mathcal L_i$ such that $d \left({x_i, \partial P}\right) = \epsilon$.

Then we have either:
 * $x_i \in \operatorname{Im} \left({\sigma}\right)$

or:
 * $x_i \in \operatorname{Im} \left({\overline \sigma}\right)$

when the Jordan curves $\sigma$ and $\overline \sigma$ gets defined from the new value of $\epsilon$.

It follows that at least two out of three of the points $q_1, q_2, q_3$ is path-connected to the same Jordan curve.

As a Jordan curve is a path, it follows that at least two of the points can be connected by a path.

Hence, $R^2 \setminus \partial P$ is the union of at most two disjoint path-connected sets.