Boundary of Polygon is Jordan Curve

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

Then there exists a Jordan curve $\gamma: \left[{0\,.\,.\,1}\right] \to \R^2$ such that the image of $\gamma$ is equal to the boundary $\partial P$ of $P$.

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 $A_{n+1} = A_1$.

As each side $S_i$ is a line segment joining $A_i$ and $A_{i+1}$, when we define the path $\gamma_i: \left[{0\,.\,.\,1}\right] \to \R^2$ by:


 * $\gamma_i \left({t}\right) = \left({1-t}\right) A_i + tA_{i+1}$

the image of $\gamma_i$ is equal to the side $S_i$.

Define $\gamma: \left[{0\,.\,.\,1}\right] \to \R^2$ as the concatenation $\left({ \cdots \left({ \left({ \gamma_1 * \gamma_2 }\right) * \gamma_3 }\right) * \ldots * \gamma_{n-1} }\right) * \gamma_n$.

Then each point in the image of $\gamma$ corresponds to a point in a side of $P$.

As $\gamma \left({0}\right) = A_1 = \gamma \left({1}\right)$ by our definition of $A_{n+1}$, it follows that $\gamma$ is a closed path.

It follows from the definition of polygon that the sides of $P$ do not intersect, except at the vertices.

For $i \ne 1$, each vertex $A_i$ is the initial point of $\gamma_i$ and the final point of $\gamma_{i-1}$, and is equal to exactly one point $\gamma \left({2^{-n - 1 + i} }\right)$ in the image of $\gamma$.

Then, $\gamma$ restricted to $ \left[{0\,.\,.\,1}\right)$ is injective.

Hence, $\gamma$ is a Jordan curve.