Paths in Interior and Exterior of Jordan Curve

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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 $p_1 \in \map {B_r}{ \map \gamma {t_1} }$ and $p_2 \in \map {B_r}{ \map \gamma {t_2} }$:

  • if $p_1, p_2 \in \Int \gamma$, then there exists a path $\sigma : \closedint 0 1 \to \Int \gamma$ with $\map \sigma 0 = p_1, \map \sigma 1 = p_2$ such that $\ds \max_{s \mathop \in \closedint 0 1} \map d {\gamma \sqbrk{\closedint {t_1}{t_2} }, \map \sigma s} < h$;
  • if $p_1, p_2 \in \Ext \gamma$, then there exists a path $\sigma : \closedint 0 1 \to \Ext \gamma$ with $\map \sigma 0 = p_1, \map \sigma 1 = p_2$ such that $\ds \max_{s \mathop \in \closedint 0 1} \map d {\gamma \sqbrk{\closedint {t_1}{t_2} }, \map \sigma s} < h$.

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 set $X$ and $y \in \R^2$.


Proof

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

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

where $\mathbb S_1$ denotes the unit circle in $\R^2$, and $\map {B_1^- }{ \mathbf 0 }$ denotes a closed ball.

For the main part of this proof, we will assume that $p_1 , p_2 \in \Int \gamma$.

We will construct a path between $\map \phi {p_1}$ and $\map \phi {p_2}$ in the interior of $\phi \circ \gamma$.

This path will be a concatenation of three segments $\sigma_{\epsilon_\sigma, \operatorname {Int} }$, $\overline \sigma_1$, and $\overline \sigma_2$.


Construction of the path segment $\sigma_{\epsilon_\sigma, \operatorname {Int} }$

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:

\(\ds \forall s \in \closedint {t_1}{t_2}: \, \) \(\ds \map { \sigma_{\epsilon, \operatorname {Int} } } s\) \(=\) \(\ds \paren { 1 - \epsilon } \paren{ \map {\phi \circ \gamma} s }\)
\(\ds \forall s \in \closedint {t_1}{t_2}: \, \) \(\ds \map { \sigma_{\epsilon, \operatorname {Ext} } } s\) \(=\) \(\ds \paren { 1 + \epsilon } \paren{ \map {\phi \circ \gamma} s }\)


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 $\map {B_1}{ \mathbf 0 }$.

From the Jordan-Schönflies Theorem, it follows that $\Img { \phi^{-1} \circ \sigma_{\epsilon, \operatorname {Int} } } \subseteq \Int \gamma$.

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 $\phi^{-1} \circ \phi \circ \gamma = \gamma$.

By definition of uniform convergence, it follows that given $h \in \R_{>0}$, we can find $\Epsilon_\sigma \in \R_{>0}$ such that for all $\epsilon_\sigma \le \Epsilon_\sigma$:

$\ds \max_{s \mathop \in \closedint {t_1}{t_2} } \norm{ \map { \phi ^{-1} \circ \sigma_{\epsilon_\sigma , \operatorname {Int} } }{ s } - \map \gamma s } < h$


Defining the radius $r$

Let $k \in \set {1, 2}$, and let $c_{ \epsilon, 1}, c _{ \epsilon, 2} : \closedint 0 1 \to \R^2$ be defined by:

$\map { c_{ \epsilon, k} }{ s } = \map {\phi \circ \gamma}{t_k} + \epsilon \tuple{ \map \cos {2 \pi s } , \map \sin {2 \pi s } }$


From Parametric Equation of Circle, it follows that $\Img { c_{ \epsilon, k} } = \map {S_\epsilon}{ \map {\phi \circ \gamma}{t_k} }$, where $\map {S_\epsilon}{ \map {\phi \circ \gamma}{t_k} }$ denotes the sphere with center $\map {\phi \circ \gamma}{t_k}$ and radius $\epsilon$.

By definition of sphere, it follows that for all $s \in \closedint 0 1$:

$\norm { \map { c_{ \epsilon, k} }{ s } - \map {\phi \circ \gamma}{t_k} } = \epsilon$


It follows that $c_{ \epsilon, k}$ converges uniformly to the constant mapping $s \mapsto \map {\phi \circ \gamma}{t_k}$, as $\epsilon$ tends to $0$.

Similar to the previous calculations, we can find $\Epsilon_1, \Epsilon_2 \in \R_{>0}$ such that for $k \in \set {1,2}$, and for all $\epsilon_k \le \Epsilon_K$:

$\ds \max_{s \mathop \in \closedint 0 1} \norm{ \map { \phi ^{-1} \circ c_{ \epsilon_k , k } }{ s } - \map \gamma {t_k} } < h$


Set $\epsilon_0 := \map \min { \Epsilon_\sigma , \Epsilon_1 , \Epsilon_2 }$.

Set $r := \ds \min_{ k \mathop \in \set {1, 2} } \max_{s \mathop \in \closedint 0 1} \norm{ \map { \phi ^{-1} \circ c_{ \epsilon_k , k } }{ s } - \map \gamma {t_k} }$.

By definition of open ball, it follows that for all $\tilde p \in \map {B_{\epsilon_0} }{ \map \gamma {t_k} }$ that $\norm {\tilde p - \map {\phi \circ \gamma}{ t_k } } < \epsilon_0$.

It follows that $\tilde p \in \map {S_{ \norm {\tilde p - \map {\phi \circ \gamma}{ t_k } } } }{ \map {\phi \circ \gamma }{t_k} }$, and $\map { \phi^{-1} }{ \tilde p } \in \map {B_r}{ \map \gamma {t_k} }$.

This implies that $\phi^{-1} \sqbrk{ \map {B_{\epsilon_0 }^- }{ \map { \phi \circ \gamma }{ t_k } } } \subseteq \map {B_r}{\map \gamma {t_k} }$.


Construction of the line segments $\overline \sigma_1$ and $\overline \sigma_2$

Let $p_1 \in \map {B_r}{ \map \gamma {t_1} }$ and $p_2 \in \map {B_r}{ \map \gamma {t_2} }$, and suppose $p_1, p_2 \in \Int \gamma$.

From the Jordan-Schönflies Theorem, it follows for $k \in \set{1, 2}$ that $\map \phi {p_k} \in \map {B_1}{ \mathbf 0 }$.

By definition of $r$, we have $\map \phi {p_k} \in \map {B_{\epsilon_0} }{ \map {\phi \circ \gamma}{ t_k } }$.

Let $\overline \sigma_1 , \overline \sigma_2 : \closedint 0 1 \to \R^2$ be the line segments between $p_k$ and $\map { \phi \circ \gamma }{ t_k }$ defined by:

$\map {\overline \sigma_1} s = s \paren{ \map { \phi \circ \gamma }{ t_1 } } + \paren { 1-s } p_1$
$\map {\overline \sigma_2} s = s p_2 + \paren { 1-s } \paren{ \map { \phi \circ \gamma }{ t_2 } }$


By definition of $\epsilon_\sigma$, we have $\map { \phi \circ \gamma }{ t_k } \in \map {B_{\epsilon_0 }^- }{ \map { \phi \circ \gamma }{ t_k } }$.

By definition of closed ball, we have $\map \phi {p_k} \in \map {B_{\epsilon_0 }^- }{ \map { \phi \circ \gamma }{ t_k } }$.

From Closed Ball is Convex Set, it follows that $\Img {\overline \sigma_k} \subseteq \map {B_{\epsilon_0 }^- }{ \map { \phi \circ \gamma }{ t_k } }$.

As we have shown that $\phi^{-1} \sqbrk{ \map {B_{\epsilon_0 }^- }{ \map { \phi \circ \gamma }{ t_k } } } \subseteq \map {B_r}{\map \gamma {t_k} }$, it follows that $\Img {\phi^{-1} \circ \overline \sigma_k } \subseteq \map {B_r}{\map \gamma {t_k} }$.

From Open Ball is Convex Set, it follows that $\Img {\overline \sigma_k} \subseteq \map {B_1}{ \mathbf 0 }$.

From the Jordan-Schönflies Theorem, it follows that $\Img {\phi^{-1} \circ \overline \sigma_k } \subseteq \Int \gamma$.


Defining the path $\sigma$

Let $\sigma: \closedint 0 1 \to \R^2$ be defined by $\sigma := \phi^{-1} \circ \paren{ \overline \sigma_1 * \sigma_{\epsilon_0 , \operatorname {Int} } * \overline \sigma_2 }$, where $*$ denotes concatenations of paths.

From what we have shown about $\Img {\overline \sigma_k}$ and $\Img { \sigma_{\epsilon_0 , \operatorname {Int} } }$, it follows that $\sigma$ is a path in $\Int \gamma$.

For all $s \in \closedint 0 1$, we have shown there exists some $s' \in \closedint {t_1}{t_2}$ such that $\norm{ \map \sigma s - \map \gamma {s'} } < h$.

By the definition of distance function, it follows that $\ds \max_{s \mathop \in \closedint 0 1} \map d {\gamma \sqbrk{\closedint {t_1}{t_2} }, \map \sigma s} < h$.


Finding a path in the exterior

Finally, suppose $p_1, p_2 \in \Ext \gamma$.

Using similar calculations, we can find a path $\sigma_{\epsilon_0 , \operatorname {Ext} }$ in $\R^2 \setminus \map {B_1^-} { \mathbf 0 }$ and two line segments $\overline \sigma_1$ and $\overline \sigma_2$ in $\map {B_{\epsilon_0 }^- }{ \map { \phi \circ \gamma }{ t_1 } }$, respectively $\map {B_{\epsilon_0 }^- }{ \map { \phi \circ \gamma }{ t_2 } }$.

As above, the path $\sigma := \phi^{-1} \circ \paren{ \overline \sigma_1 * \sigma_{\epsilon_0 , \operatorname {Ext} } * \overline \sigma_2 }$ will fulfil the requirements.

$\blacksquare$

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