Boundary of Polygon is Topological Boundary

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Theorem

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

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

Let $\Int P$ and $\Ext P$ denote the interior and exterior of $\partial P$, where $\partial P$ is considered as a Jordan curve.


Then the topological boundary of $\Int P$ is equal to $\partial P$, and the topological boundary of $\Ext P$ is equal to $\partial P$.


Proof

Denote the topological boundary of $\Int P$ as $\partial \Int P$, and denote the topological boundary of $\Ext P$ as $\partial \Ext P$.


Topological Boundary is Subset of Boundary

From Boundary of Polygon is Jordan Curve, it follows that the boundary $\partial P$ is equal to the image of a Jordan curve.

From Jordan Polygon Theorem, it follows that $\Int P$ and $\Ext P$ are disjoint, open and path-connected.

From Set is Open iff Disjoint from Boundary, it follows that $\Int P$ and $\partial \Int P$ are disjoint.

From Disjoint Open Sets remain Disjoint with one Closure, it follows that $\Ext P$ and the closure of $\Int P$ are disjoint.

As $\partial \Int P$ is a subset of the closure of $\Int P$, it follows that $\Ext P$ and $\partial \Int P$ are disjoint.

As $\R^2 = \Int P \cup \Ext P \cup \partial P$ by the Jordan Polygon Theorem, it follows that $\partial \Int P \subseteq \partial P$.

Similarly, it follows that $\partial \Ext P \subseteq \partial P$.

$\Box$


Boundary is Subset of Topological Boundary

Let $p \in \partial P$ such that $p$ is not a vertex, and let $S$ be the side of $P$ that $p$ is a part of.

Denote the $j$th side of $P$ as $S_j$, and let $n \in \N$ be the total number of sides.

Let $\displaystyle \delta = \map d {S, \bigcup_{j \mathop = 1, \ldots, n: S_j \mathop \ne S} S_j}$ be the Euclidean distance between $S$ and all other sides of $P$.

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

Let $\epsilon \in \openint 0 \delta$, and denote the open ball of $p$ with radius $\epsilon$ as $\map {B_\epsilon} p$.

Choose $x_1 \in \map {B_\epsilon} p$, and put $\mathbf v = p - x_1$.

Let $\LL_1 = \set {x_1 + s \mathbf v: s \in \R_{\ge 0} }$ be a ray with start point $x_1$.

Then $\LL_1$ and $S$ has one crossing at $p$.

Put $x_2 = x_1 + 2 \mathbf v$, and put $\LL_2 = \set {x_2 + s \mathbf v: s \in \R_{\ge 0} }$, so $\LL_1 \cap \LL_2 = \LL_2$.

Then $\LL_2$ and $S$ do not cross.

As $x_2 \in \map {B_\epsilon} p$ with $\epsilon < \delta$, it follows from the definition of $\delta$ that if $\LL_1$ and some side $S'$ has a crossing, then $\LL_2$ and $S'$ also has a crossing.

If $\map N {x_i}$ denotes the number of crossings between $\LL_i$ and $\partial P$, it follows that $\map N {x_1} + 1 = \map N {x_2}$.

Then $\map {\mathrm {par} } {x_1} \ne \map {\mathrm {par} } {x_2}$, where $\map {\mathrm {par} } {x_i}$ denotes the parity of $x_i$.

From Jordan Polygon Interior and Exterior Criterion, it follows that one of the points $x_1, x_2$ belongs to $\Int P$, and the other point belongs to $\Ext P$.

As $\epsilon$ was arbitrary small, it follows that $p$ is a limit point of both $\Int P$ and $\Ext P$.

By definition of closure, it follows that $p$ lies in the closure of $\Int P$ and $\Ext P$.

Then $p \in \partial \Int P$ and $p \in \partial \Ext P$, as the Jordan Polygon Theorem shows that $\partial P$ and $\Int P$, $\Ext P$ are disjoint.


Now, suppose that $p$ is a vertex of $S$.

Then we can find a sequence $\sequence {p_k}$ of points that lies on the adjacent sides of $p$ such that the sequence converges to $p$.

As none of the point in $\sequence {x_k}$ are vertices, all $x_k$ lie in $\partial \Int P$ and $\partial \Ext P$.

As Boundary of Set is Closed, it follows that $p \in \partial \Int P$, and $p \in \partial \Ext P$.

Hence, $\partial P \subseteq \partial \Int P$, and $\partial P \subseteq \partial \Ext P$.

$\Box$


The result now follows by definition of set equality.

$\blacksquare$