Jordan Curve and Jordan Arc form Two Jordan Curves

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Theorem

Let $\left[{a \,.\,.\, b}\right]$ denote the closed real interval between $a \in \R, b \in \R: a \le b$.

Let $\gamma: \left[{a \,.\,.\, b}\right] \to \R^2$ be a Jordan curve.

Let the interior of $\gamma$ be denoted $\operatorname{Int} \left({\gamma}\right)$.

Let the image of $\gamma$ be denoted $\operatorname{Im} \left({\gamma}\right)$.


Let $\sigma: \left[{c \,.\,.\, d}\right] \to \R^2$ be a Jordan arc such that:

$\sigma \left({c}\right) \ne \sigma \left({d}\right)$
$\sigma \left({c}\right), \sigma \left({d}\right) \in \operatorname{Im} \left({\gamma}\right)$

and:

$\forall t \in \left({c \,.\,.\, d}\right): \sigma \left({t}\right) \in \operatorname{Int} \left({\gamma}\right)$


Let $t_1 = \gamma^{-1} \left({ \sigma \left({c}\right) }\right)$.

Let $t_2 = \gamma^{-1} \left({ \sigma \left({d}\right) }\right)$.

Let $t_1 < t_2$.


Define:

$-\sigma: \left[{c \,.\,.\, d}\right] \to \operatorname{Im} \left({\sigma}\right)$ by $-\sigma \left({t}\right) = \sigma \left({c + d - t}\right)$

Let $*$ denote concatenation of paths.

Let $\gamma \restriction_{\left[{a \,.\,.\, t_1}\right] }$ denote the restriction of $\gamma$ to $\left[{a \,.\,.\, t_1}\right]$.


Define:

$\gamma_1 = \gamma {\restriction_{\left[{a \,.\,.\, t_1}\right] } } * \sigma * \gamma{\restriction_{\left[{t_2 \,.\,.\, b}\right] } }$

Define:

$\gamma_2 = \gamma {\restriction_{\left[{t_1 \,.\,.\, t_2}\right] } } * \left({ -\sigma }\right)$


Then $\gamma_1$ and $\gamma_2$ are Jordan curves such that:

$\operatorname{Int} \left({\gamma_1}\right) \subseteq \operatorname{Int} \left({\gamma}\right)$

and:

$\operatorname{Int} \left({\gamma_2}\right) \subseteq \operatorname{Int} \left({\gamma}\right)$


Corollary

Define:

$\gamma_1 = \gamma {\restriction_{\closedint a {t_2} } } * \paren {-\sigma} * \gamma {\restriction_{\closedint {t_1} b} }$

Define:

$\gamma_2 = \gamma {\restriction_{\closedint {t_2} {t_1} } } * \sigma$


Then $\gamma_1$ and $\gamma_2$ are Jordan curves such that:

$\map {\operatorname {Int} } {\gamma_1} \subseteq \map {\operatorname {Int} } \gamma$

and:

$\map {\operatorname {Int} } {\gamma_2} \subseteq \map {\operatorname {Int} } \gamma$


Proof

As:

$\operatorname{Int} \left({\gamma}\right)$ and $\operatorname{Im} \left({\gamma}\right)$ are disjoint by the Jordan Curve Theorem

and:

$\sigma \left({ \left({c \,.\,.\, d}\right) }\right) \subseteq \operatorname{Int} \left({\gamma}\right)$

it follows that:

$\operatorname{Im} \left({\gamma}\right) \cap \operatorname{Im} \left({\sigma}\right) = \left\{ { \sigma \left({c}\right), \sigma \left({d}\right) }\right\}$.

As $\gamma$ is a Jordan curve, it follows that $\gamma {\restriction_{ \left[{a \,.\,.\, t_1}\right] } }$ and $\gamma{ \restriction_{ \left[{t_2 \,.\,.\, b}\right] } }$ intersect only in $\gamma \left({a}\right)$.

It follows that $\gamma_1$ is a Jordan arc.


As the initial point of $\gamma_1$ is $\gamma \left({a}\right)$, and the final point of $\gamma_1$ is $\gamma \left({b}\right) = \gamma \left({a}\right)$, it follows that $\gamma_1$ is a Jordan curve.

As $\operatorname{Im} \left({-\sigma}\right) = \operatorname{Im} \left({\sigma}\right)$, it follows that $\gamma_2$ is a Jordan arc.

As $\gamma \left({t_1}\right) = \sigma \left({c}\right) = -\sigma \left({d}\right)$, it follows that $\gamma_2$ is a Jordan curve.

$\Box$


Denote the exterior of $\gamma$ as $\operatorname{Ext} \left({\gamma}\right)$.

Let $q_0 \in \operatorname{Ext} \left({\gamma}\right)$ be determined.

Let $q \in \operatorname{Ext} \left({\gamma}\right)$.

By the Jordan Curve Theorem, $\operatorname{Ext} \left({\gamma}\right)$ is unbounded.

Hence for all $N \in \N$ we can choose $q \in \operatorname{Ext} \left({\gamma}\right)$ such that:

$d \left({\mathbf 0, q}\right) > N$

where $d$ denotes the Euclidean metric on $\R^2$.



The Jordan Curve Theorem also shows that $\operatorname{Ext} \left({\gamma}\right)$ is open and connected.

From Connected Open Subset of Euclidean Space is Path-Connected, there exists a path:

$\rho: \left[{0 \,.\,.\, 1}\right] \to \operatorname{Ext} \left({\gamma}\right)$

joining $q$ and $q_0$.


We have:

$\operatorname{Im} \left({\gamma_1}\right) \subseteq \operatorname{Int} \left({\gamma}\right) \cup \operatorname{Im} \left({\gamma}\right)$

This is disjoint with $\operatorname{Ext} \left({\gamma}\right)$

Thus it follows that $\rho$ is a path in either $\operatorname{Ext} \left({\gamma_1}\right)$ or $\operatorname{Int} \left({\gamma_1}\right)$.


We have that $d \left({\mathbf 0, q}\right)$ can be arbitrary large.

Also, $\operatorname{Int} \left({\gamma_1}\right)$ is bounded.

So follows that $\rho$ is a path in $\operatorname{Ext} \left({\gamma_1}\right)$.

In particularly:

$q \in \operatorname{Ext} \left({\gamma_1}\right)$

Therefore:

$\operatorname{Ext} \left({\gamma}\right) \subseteq \operatorname{Ext} \left({\gamma_1}\right)$


Let $q_1 \in \operatorname{Int} \left({\gamma_1}\right)$.

Then:

$q_1 \notin \operatorname{Ext} \left({\gamma}\right)$

as $\operatorname{Int} \left({\gamma_1}\right)$ and $\operatorname{Ext} \left({\gamma_1}\right)$ are disjoint.

It follows that:

$q_1 \in \operatorname{Int} \left({\gamma_1}\right)$

Therefore:

$\operatorname{Int} \left({\gamma_1}\right) \subseteq \operatorname{Int} \left({\gamma}\right)$

Similarly, it follows that:

$\operatorname{Int} \left({\gamma_2}\right) \subseteq \operatorname{Int} \left({\gamma}\right)$

$\blacksquare$