User:Anghel/Sandbox

Theorem
Let $\gamma : \closedint 0 1 \to \R^2$ be a Jordan curve.

Let $t_1, t_2 \in \closedint 0 1$ with $t_1 < t_2$.

Let $h \in \R_{>0}$.

Let $\Img {\gamma}$, $\Int { \gamma }$ and $\Ext { \gamma }$ denote the image, interior, and exterior of $\gamma$.

Then there exists $r \in \R_{>0}$ such that $r \le h$, and for all $p \in \map {B_r}{ \map \gamma {t_1} }, q \in \map {B_r}{ \map \gamma {t_2} }$:


 * if $p, q \in \Int \gamma$, then there exists a path $\sigma : \closedint 0 1 \to \Int \gamma$ with $\map \sigma 0 = p, \map \sigma 1 = q$ such that $\map d {\Img \sigma, \Img \gamma} < h$;


 * if $p, q \in \Ext \gamma$, then there exists a path $\sigma : \closedint 0 1 \to \Ext \gamma$ with $\map \sigma 0 = p, \map \sigma 1 = q$ such that $\map d {\Img \sigma, \Img \gamma} < h$.

Here $\map {B_r}{ x }$ denotes the open ball with radius $r$ and center $x$, and $\map d {X,Y}$ denotes the distance between the sets $X$ and $Y$.

Proof
Let $\phi : \R^2 \to \R^2$ be the homeomorphism defined in the Jordan-Schönflies Theorem such that:


 * $\phi \sqbrk {\Img {\gamma} } = \mathbb S_1$
 * $\phi \sqbrk {\Int {\gamma} } = \map {B_1}{ \mathbf 0 }$
 * $\phi \sqbrk {\Ext {\gamma} } = \R^2 \setminus \map {B_1^-} { \mathbf 0 }$

where $\mathbb S_1$ denotes the unit circle in $\R^2$.

For all $\epsilon \in \openint 0 {\dfrac 1 2}$, let $\sigma_{\epsilon, \operatorname {Int} }, \sigma_{\epsilon, \operatorname {Ext} } : \closedint {t_1}{t_2}$ be the paths defined by:


 * for all $s \in \closedint {t_1}{t_2}$: $\map { \sigma_{\epsilon, \operatorname {Int} } } s = \paren { 1 - \epsilon } \paren{ \map {\phi \circ \gamma} s }$
 * for all $s \in \closedint {t_1}{t_2}$: $\map { \sigma_{\epsilon, \operatorname {Ext} } } s = \paren { 1 + \epsilon } \paren{ \map {\phi \circ \gamma} s }$

As $\norm{ \map {\phi \circ \gamma} s } = 1$, it follows that $\norm{ \map { \sigma_{\epsilon, \operatorname {Int} } } s } = 1 - \epsilon$, where $\norm {\mathbf x}$ denotes the Euclidean norm of $\mathbf x$ on $\R^2$.

It follows that $\sigma_{\epsilon, \operatorname {Int} }$ is a path in $\map {B_1}{ \mathbf 0 }$.

For all $s \in \closedint 0 1$, the distance between $\map { \sigma_{\epsilon, \operatorname {Int} } } s$ and $\map {\phi \circ \gamma} s$ is:


 * $\norm { \map { \sigma_{\epsilon, \operatorname {Int} } } s - \map {\phi \circ \gamma} s} = \epsilon \norm {\map  {\phi \circ \gamma} s} = \epsilon$

which is independent of the value of $s$, and converges to $0$ as $\epsilon$ tends to $0$.

It follows that $\sigma_{\epsilon, \operatorname {Int} }$ converges to $\phi \circ \gamma$ uniformly as $\epsilon$ tends to $0$.

Closed Real Interval is Compact shows that $\closedint 0 1$ is compact.

From Composition of Continuous Mapping on Compact Space Preserves Uniform Convergence, it follows that $\phi ^{-1} \circ \sigma_{\epsilon, \operatorname {Int} }$ converges to $\phi^{-1} \circ \phi \circ \gamma$ uniformly as $\epsilon$ tends to $0$.

From Composite of Bijection with Inverse is Identity Mapping, it follows that $\phi^{-1} \circ \phi \circ \gamma = \gamma$.

It follows that given $h \in \R_{>0}$, we can find $\epsilon_0 \in \R_{>0}$ such that:


 * $\map d { \Img { \phi ^{-1} \circ \sigma_{\epsilon, \operatorname {Int} } }, \Img \gamma } < h$

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

A similar argument shows that we can find $\delta_0 \in \R_{>0}$ such that:


 * $\map d { \Img { \phi ^{-1} \circ \sigma_{\delta_0, \operatorname {Ext} } }, \Img \gamma } < h$

The set $\map {B_{\epsilon } }{ \map { \phi \circ \gamma_0 }{ t_1 } } \setminus \mathbb S^1$ is split into these two path components:


 * $B_{\operatorname{Int} } = \map {B_{\epsilon } }{ \map { \phi \circ \gamma_0 }{ t_1 } } \cap \map {B_1}{ \mathbf 0 }$
 * $B_{\operatorname{Ext} } = \set { p' \in \map { B_{\epsilon } }{ \map { \phi \circ \gamma_0 }{ t_1 } } : \norm {p'} > 1 }$

Set $U_1 := \phi^{-1} \sqbrk{ B_{\operatorname{Int} } }$.

Set $U_2 := \phi^{-1} \sqbrk{ \map {B_{\epsilon } }{ \map { \phi \circ \gamma_0 }{ t_2 } } \cap \map {B_1}{ \mathbf 0 } }$.

From Finite Intersection of Open Sets of Normed Vector Space is Open and Mapping is Continuous iff Inverse Images of Open Sets are Open, it follows that $U_1$ and $U_2$ are open in $\R^2$.

Let $\tilde p \in B_{\operatorname{Int} }$.

By definition of $\sigma_{\epsilon_0, \operatorname {Int} }$, we have $\map { \sigma_{\epsilon, \operatorname {Int} } }{ 0 } \in B_{\operatorname{Int} }$.

Let ${l_1} : \closedint 0 1 \to \R^2$ be the line segment between $\tilde p$ and $\map { \sigma_{\epsilon, \operatorname {Int} } }{ 0 } $ defined by:


 * $\map {l_1} s = s \map { \sigma_{\epsilon, \operatorname {Int} } }{ 0 } + \paren { 1-s } \tilde p$

From Open Ball is Convex Set, it follows that $\Img {l_1} \subseteq B_{\operatorname{Int} }$.

As $ Category:Jordan Curves]]