# Jordan Curve and Jordan Arc form Two Jordan Curves

## Theorem

Let $\closedint a b$ denote the closed real interval between $a \in \R, b \in \R: a \le b$.

Let $\gamma: \closedint a b \to \R^2$ be a Jordan curve.

Let the interior of $\gamma$ be denoted $\Int \gamma$.

Let the image of $\gamma$ be denoted $\Img \gamma$.

Let $\sigma: \closedint c d \to \R^2$ be a Jordan arc such that:

$\map \sigma c \ne \map \sigma d$
$\map \sigma c, \map \sigma d \in \Img \gamma$

and:

$\forall t \in \openint c d: \map \sigma t \in \Int \gamma$

Let $t_1 = \map {\gamma^{-1} } {\map \sigma c}$.

Let $t_2 = \map {\gamma^{-1} } {\map \sigma d}$.

Let $t_1 < t_2$.

Define:

$-\sigma: \closedint c d \to \Img \sigma$ by $-\map \sigma t = \map \sigma {c + d - t}$

Let $*$ denote concatenation of paths.

Let $\gamma \restriction_{\closedint a {t_1} }$ denote the restriction of $\gamma$ to $\closedint a {t_1}$.

Define:

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

Define:

$\gamma_2 = \gamma {\closedint a {t_1} } * \paren {-\sigma}$

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

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

and:

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

### 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:

$\Int \gamma$ and $\Img \gamma$ are disjoint by the Jordan Curve Theorem

and:

$\map \sigma {\openint c d} \subseteq \Int \gamma$

it follows that:

$\Img \gamma \cap \Img \sigma = \set {\map \sigma c, \map \sigma d}$.

As $\gamma$ is a Jordan curve, it follows that $\gamma {\restriction_{\closedint a {t_1} } }$ and $\gamma {\restriction_{\closedint {t_2} b} }$ intersect only in $\map \gamma a$.

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

As the initial point of $\gamma_1$ is $\map \gamma a$, and the final point of $\gamma_1$ is $\map \gamma b = \map \gamma a$, it follows that $\gamma_1$ is a Jordan curve.

As $\Img {-\sigma} = \Img \sigma$, it follows that $\gamma_2$ is a Jordan arc.

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

$\Box$

Denote the exterior of $\gamma$ as $\Ext \gamma$.

Let $q_0 \in \Ext \gamma$ be determined.

Let $q \in \Ext \gamma$.

By the Jordan Curve Theorem, $\Ext \gamma$ is unbounded.

Hence for all $N \in \N$ we can choose $q \in \Ext \gamma$ such that:

$\map d {\mathbf 0, q} > N$

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

The Jordan Curve Theorem also shows that $\Ext \gamma$ is open and connected.

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

$\rho: \closedint 0 1 \to \Ext \gamma$

joining $q$ and $q_0$.

We have:

$\Img {\gamma_1} \subseteq \Int \gamma \cup \Img \gamma$

This is disjoint with $\Ext \gamma$

Thus it follows that $\rho$ is a path in either $\Ext {\gamma_1}$ or $\Int {\gamma_1}$.

We have that $\map d {\mathbf 0, q}$ can be arbitrary large.

Also, $\Int {\gamma_1}$ is bounded.

So follows that $\rho$ is a path in $\Ext {\gamma_1}$.

In particularly:

$q \in \Ext {\gamma_1}$

Therefore:

$\Ext \gamma \subseteq \Ext {\gamma_1}$

Let $q_1 \in \Int {\gamma_1}$.

Then:

$q_1 \notin \Ext \gamma$

as $\Int {\gamma_1}$ and $\Ext {\gamma_1}$ are disjoint.

It follows that:

$q_1 \in \Int {\gamma_1}$

Therefore:

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

Similarly, it follows that:

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

$\blacksquare$