I've been trying to find a way to define a simple polygon with sets of points and segments. I have some questions about this, but I need to introduce a few things first.
Let $n\in\{3,4,\dots\}$ be some finite constant, and $I=\{1,2,\dots,n\}$. Then let $p_i\in\Bbb R^2$ for all $i\in I$. Assume that $\forall i,j\in I, p_i\neq p_j\iff i\neq j$. Then let $p=\{p_i:i\in I\}$. Finally, assume that there is no line $l$, where $p\subset l$.
Then, we define, for $a,b\in\Bbb R^2$, a segment function $$s(a,b)=\{a+t(b-a):t\in[0,1]\}$$ With this, we define, for any $i,j\in I$, $$s_{ij}=s(p_i,p_j)$$ Then we define $$\phi=\{s_{ij}:(i,j)\in I\times I\}$$ Therefore, $$p\subset\phi$$
My question:
How do we rigorously define the closed, simple path made out of segments, $\gamma\subseteq\phi$, where $p\subset\gamma$?
Update: I found an answer! Check out my answer below.
Update 2.0: I didn't actually find an answer! But this is how far I've come:
First of all, it is incorrect that $p\subset\phi$, because $p\neq\{\{p_i\}:i\in I\}$. I had noticed that $\{\{p_i\}:i\in I\}\subset\phi$, because $\{\{p_i\}:i\in I\}=\{s_{ii}:i\in I\}\subset\phi$. Therefore $p\cap\phi=\emptyset$.
Next, I noticed that, if we define $$\Phi=\phi\setminus\{\{p_i\}:i\in I\}$$ and we draw out some diagrams of $\Phi$, we notice something interesting. If $A\in\Phi$, and $A$ intersects any other segments in $\Phi$, then $A\not\in\gamma$. The remaining task is to find a set $G\subseteq\Phi$ where all elements of $G$ do not intersect any elements of $\Phi$ except for themselves. A possible definition: $$G=\{A\in\Phi:\forall B\in\Phi, (A\cap B)\in\{A,\emptyset\}\}$$ Where $\emptyset=\{\}$ is the empty set (for clarification).
In any case, given the definition (that I gave in words) for $G$, $$\gamma=G\cup p$$