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 $\displaystyle \delta_i = \map d {S_i, \bigcup_{j \mathop = 1, \ldots, n: \, \size {i - j} > 1 } S_j}$ 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: :$\gamma: \closedint 0 1 \to \R^2$ 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 $\mathbf v_i = \dfrac {A_{i + 1} - A_i} {\map d {A_{i + 1}, A_i} }$ be the direction vector of $\gamma_i$ with norm $\norm {\mathbf v_i} = 1$. Let $\mathbf w_i$ be the vector $v_i$ rotated $\dfrac \pi 2$ radians counterclockwise, so $\norm {\mathbf w_i} = 1$. {{explain|Invoke the fact that rotation is an isometry (still to be covered on ).}} For any $\epsilon \in \openint 0 {\dfrac \delta 2}$, we intend to construct two Jordan curves $\sigma, \overline \sigma$ such that: :$\Img \sigma \cup \Img {\overline \sigma} = \set {q \in \R^2: \map d {q, \partial P} = \epsilon}$ rightthumbIllustration of the [[Definition:Jordan CurveJordan curves $\sigma, \overline \sigma$ which both have distance $\epsilon$ to $\partial P$.<br/>Note the rounded corners of the curves that occur in place of non-convex angles.]] For $i \in \set {1, \ldots, n}$, 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 $\map {\rho_i} t = p_{i + 1}$. 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 $\sigma: \closedint 0 1 \to \R^2$ 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 $i, j \in \set {1, \ldots, n}$, $\sigma_i$ 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 $\size {i - j} > 1$: :let $x_i \in \Img {\sigma_i} \cup \Img {\rho_i}$ :let $x_j \in \Img {\sigma_j} \cup \Img {\rho_j}$. Let $p_i, p_j \in \partial P$ be two points such that: :$\map d {x_i, p_i} = \map d {x_j, p_j} = \epsilon$ Suppose $x_i \in \Img {\sigma_i}$. Then we can put: :$p_i = x_i - \epsilon \mathbf w_i$ Suppose $x_i \in \Img {\rho_i}$. Then we can put $p_i = A_{i + 1}$. Then $x_i \ne x_j$, as: {{begin-eqn}} {{eqn | l = \map d {x_i, x_j}       | r = \map d {x_i, x_j} + \map d {x_i, p_i} + \map d {x_j, p_j} - 2 \epsilon }} {{eqn | o = \ge       | r = \map d {p_i, p_j} - 2 \epsilon       | c = Triangle Inequality for Vectors in Euclidean Space }} {{eqn | o = >       | r = \delta - 2 \epsilon       | c = as $p_i \in S_i, p_j \in S_j$, and $\map d {S_i, S_j} \ge \delta$ }} {{eqn | r = 0 }} {{end-eqn}} 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 $i \in \set {1, \ldots, n}$, 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 $i, j \in \set {1, \ldots, n}$, $\sigma_i$ and $\rho_i$ do not intersect with $\overline \sigma_j$ and $\overline \rho_j$.  This follows as: :$\map d {\sigma_i, S_i} = \map d {\rho_i, S_i} = \map d {\rho_i, S_{i - 1} } = \epsilon$ and as $S_i$, or $S_{i - 1}$, are not adjacent sides to $S_j$, we have $\map d {S_i, S_j} > 2 \epsilon$. Now we can use the Triangle Inequality for Vectors in Euclidean Space as above, to prove that the paths do not intersect. {{qed|lemma}} Finally, let $q_1, q_2, q_3 \in R^2 \setminus \partial P$.  Let: :$\displaystyle \epsilon = \min \set {\map d {q_1, \partial P} / 2, \map d {q_2, \partial P} / 2, \map d {q_3, \partial P} / 2, \delta}$ For all $i \in \set {1, 2, 3}$, draw a line segment $\LL_i$ joining $q_i$ with any point on the boundary $\partial P$. As $\map d {q_i, \partial P} > \epsilon$, it follows from the Intermediate Value Theorem that there is a point $x_i$ on $\LL_i$ such that $\map d {x_i, \partial P} = \epsilon$. Then we have either: :$x_i \in \Img \sigma$ or: :$x_i \in \Img {\overline \sigma}$ 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.  == Sources == {{SourceReview|Revisit this link}} [http://compgeom.cs.uiuc.edu/~jeffe/teaching/comptop/2009/notes/jordan-polygon-theorem.pdf Jeff Ericsson: Computational Topology]