User:Anghel/Sandbox

Theorem
Let $\gamma_0, \gamma_1 , \gamma_2 : \closedint 0 1 \to \R^2$ be Jordan arcs such that:


 * $ \map {\gamma_0} 0 = \map {\gamma_1} 1 = \map {\gamma_2} 1$, $\map {\gamma_0} 1 = \map {\gamma_1} 0 = \map {\gamma_2} 0$

Let $\gamma_0 \sqbrk { \openint 0 1 }, \gamma_1 \sqbrk { \openint 0 1 }, \gamma_2 \sqbrk { \openint 0 1 }$ be pairwise disjoint.

Let $\mathbf c \in \R^2$ and $r_0 \in \R_{>0}$ such that:


 * $\Img {\gamma_1} \cup \Img {\gamma_2} \subseteq \map {B_{r_0} }{ \mathbf c }$

where $\Img {\gamma_1}$ denotes the image of $\gamma_1$, and $\map {B_{r_0} }{ \mathbf c }$ denotes the open ball with radius $r_0$ and center $\mathbf c$.

Let $t \in \openint 0 1$ such that $\map {\gamma_0} t \notin \map {B_{r_0} }{ \mathbf c }$.

Then there exists $r \in \R_{>0}$ such that for all $p \in \map {B_r}{ \map {\gamma_0} t }$:


 * $p \in \Int { \gamma_0 * \gamma_1 }$, $p \in \Int { \gamma_0 * \gamma_2 }$
 * $p \in \Ext { \gamma_0 * \gamma_1 }$, $p \in \Ext { \gamma_0 * \gamma_2 }$

where $\Int { \gamma_0 * \gamma_1 }$ and $\Ext { \gamma_0 * \gamma_1 }$ denote the interior, respectively the exterior of the Jordan curve created by concatenating $\gamma_0$ and $\gamma_1$.

Proof
By definition of Jordan curve, it follows that $\gamma_0 * \gamma_1$ and $\gamma_0 * \gamma_2$ are Jordan curves.

Let $l : \hointr 0 \infty \to \R^2$ be the ray with start point $\mathbf c$ that passes through $\map {\gamma_0} t$ defined by:


 * $\map l s = \mathbf c + s \paren { \map {\gamma_0} t - \mathbf c }$

Set $s_0 = \ds \sup_{ s \in \hointr 0 \infty } \set { \map l s \in \Img { \gamma_0 } }$.

From Continuous Image of Compact Space is Compact, it follows that $\Img { \gamma_0 }$ is compact.

By definition of compact sets, it follows that $\Img { \gamma_0 }$ is bounded, so $s_0 \in \R$.

By definition of Jordan arc, it follows that $\gamma_0$ is injective.

As $s_0 \in \Img { \gamma_0 }$, it follows from Injection to Image is Bijection that $t_0 := \map { \gamma_0^{-1} }{ s_0 }$ is well-defined.

Let $l \restriction_{ \openint {s_0} \infty }$ denote the restriction of $l$ to $\openint {s_0} \infty$.

By definition of $s_0$, it follows that $l \restriction_{ \openint {s_0} \infty }$ does not intersect $\gamma_0$.

As for all $s \in \openint {s_0} \infty $, we have:

so by assumption of $\Img {\gamma_1} \cup \Img {\gamma_2} \subseteq \map {B_{r_0} }{ \mathbf c }$, it follows that $l \restriction_{ \openint {s_0} \infty }$ does not intersect $\gamma_1$ or $\gamma_2$.

As $\Img { l \restriction_{ \openint {s_0} \infty } }$ is unbounded, it follows by definition of exterior that $\map l s \in \Ext {\gamma_0 * \gamma_1} \cup \Ext {\gamma_0 * \gamma_2}$ for all $s > s_0$.

Let $\mathbb I \subseteq \closedint 0 1$ be the closed real interval with endpoints $\tilde t, \tilde t_0$ defined such that $\map {\gamma_0 * \gamma_1} {\tilde t} = \map {\gamma_0} t, \map {\gamma_0 * \gamma_1} {\tilde t_0} = \map {\gamma_0}{ t_0 }$.

By definition of $\gamma_0, \gamma_1, \gamma_2$, it follows that $\gamma_0 * \gamma_1 \sqbrk { \mathbb I }$ is disjoint with $\Img {\gamma_1} \cup \Img {\gamma_2}$.

Note that for all $s \in \mathbb I$, we have $\map {\gamma_0 * \gamma_1} s = \map {\gamma_0 * \gamma_2} s$.

From Continuous Image of Compact Space is Compact, it follows that $\gamma_0 * \gamma_1 \sqbrk{ \mathbb I }$ and $\Img {\gamma_1} \cup \Img {\gamma_2}$ are compact.

From Distance between Disjoint Compact Set and Closed Set in Metric Space is Positive, it follows that:


 * $\map d {\gamma_0 * \gamma_1 \sqbrk{ \mathbb I }, \Img {\gamma_1} \cup \Img {\gamma_2} } > 0$

where $\map d {X, Y}$ denotes the distance between the sets $X$ and $Y$ induced by the Euclidean norm on $\R^2$.

Set $h := \map d {\gamma_0 * \gamma_1 \sqbrk{ \mathbb I }, \Img {\gamma_1} \cup \Img {\gamma_2} } / 2$.