User:Anghel/Sandbox

Proof
Set $l := \begin {cases} k+1 & : k < N \\ 1 & : k = N \end {cases}$.

From Continuous Image of Compact Space is Compact and Finite Union of Compact Sets is Compact, it follows that


 * $\ds \bigcup_{ n \mathop = 1 }^{k-1} \Img {\gamma_n} \cup \bigcup_{n \mathop = l+1}^{N} \Img {\gamma_n}$

is compact.

Set $r_0 := \ds \map d {\bigcup_{ n \mathop = 1 }^{k-1} \Img {\gamma_n} \cup \bigcup_{n \mathop = l + 1}^{N} \Img {\gamma_n}, \map {\gamma_k}{c_k} } / 2$.

From Distance between Disjoint Compact Set and Closed Set in Metric Space is Positive, it follows that $r_0 > 0$.

Set:


 * $\ds \tilde t_1 := \max \set {t \mathop \in \closedint {c_{k-1} }{ c_k } : \map {\gamma_k} t \in \map { N_{r_0}^- }{ \map {\gamma_k}{c_k} } }$
 * $\ds \tilde t_2 := \min \set {t \mathop \in \closedint {c_{l-1} }{ c_l } : \map {\gamma_l} t \in \map { N_{r_0}^- }{ \map {\gamma_k}{c_k} } }$

where $\map { N_{r_0}^- }{ \map {\gamma_k}{c_k} }$ denotes a closed disk.

Set $r_1 := \map d { \gamma_k \sqbrk{ \closedint {c_{k-1} }{\tilde t_1} } \cup \gamma_l \sqbrk{ \closedint {\tilde t_2}{c_l} }, \map {\gamma_k}{c_k} } / 2$.

From Distance between Disjoint Compact Set and Closed Set in Metric Space is Positive, it follows that $r_1 > 0$.

Set :


 * $\tilde t_3 := \min \set{t \in \closedint {\tilde t_1}{c_k} : \exists \delta > 0 : \forall \epsilon \in \openint 0 \delta : \map {\gamma_k}{t + \epsilon} \in \map { N_{r_1} }{ \map {\gamma_k}{c_k} } }$
 * $\tilde t_4 := \max \set{t \in \closedint {c_{l-1} }{\tilde t_2} : \exists \delta > 0 : \forall \epsilon \in \openint 0 \delta : \map {\gamma_k}{t - \epsilon} \in \map { N_{r_1} }{ \map {\gamma_k}{c_k} } }$

where $\map { N_{r_1}^- }{ \map {\gamma_k}{c_k} }$ denotes an open disk.

As $r_1 < r_0$, this implies that:

where $n \notin \set {k, l}$.

Lemma
Let $\tilde \gamma_1 : \closedint a b \to \C$ be an injective piecewise continuously differentiable function such that $\map {\tilde \gamma_1 '}{t} \ne 0$ for all $t \in \openint a b$.

Let $\tilde t_3, \tilde t_4 \in \openint a b$ with $\tilde t_3 < \tilde t_4$.

Let $\tilde t_5 \in \openint {\tilde t_3}{\tilde t_4}$ such that $\tilde \gamma_1$ is complex-differentiable at all $t \in \openint a b \setminus \set {\map{ \tilde \gamma_1 }{ \tilde t_5 } }$.

Let $r_1 \in \R_{>0}$ such that:


 * $r_1 = \cmod{ \map {\tilde \gamma_1}{\tilde t_3} - \map {\tilde \gamma_1}{\tilde t_5} } = \cmod { \map {\tilde \gamma_1}{\tilde t_4} - \map {\tilde \gamma_1}{\tilde t_5} }$

Let $\delta \in \R_{>0}$ such that for all $\epsilon \in \openint 0 \delta$:


 * $\map {\tilde \gamma_1}{\tilde t_3 + \epsilon}, \map {\tilde \gamma_1}{\tilde t_4 - \epsilon} \in \map { N_{r_1} }{ \map {\tilde \gamma_1}{\tilde t_5} }$

where $\map { N_{r_1} }{ \map {\tilde \gamma_1}{\tilde t_5} }$ denotes an open disk.

Then there exists an injective complex-differentiable function $\sigma : \closedint a b \to \C$ such that:


 * $\map \sigma t = \map {\tilde \gamma_1}{t}$ for all $t \in \closedint {a}{\tilde t_3} \cup \closedint {\tilde t_4}{b}$
 * $\map \sigma t \in \map {N_{r_1} }{ \map {\tilde \gamma_1}{\tilde t_5} }$ for all $t \in \openint{\tilde t_3}{\tilde t_4}$

From Complex Plane is Homeomorphic to Real Plane, it follows that the complex plane $\C$ can be identified with the Euclidean plane $\R^2$ by the homeomorphism $\phi : \R^2 \to \C$, defined by $\map \phi {x, y} = x + i y$.

Define two Jordan arcs $\tilde \gamma_0, \tilde \gamma_1 : \closedint 0 1 \to \C$ by:


 * $\tilde \gamma_0 := \gamma_l {\restriction_{\closedint {\tilde t_4}{c_l} } } * \gamma_{l+1} * \gamma_{l+2} * \ldots * \gamma_n * \gamma_1 * \gamma_2 * \ldots * \gamma_{k-1} * \gamma_k {\restriction{ \closedint {c_{k-1} }{\tilde t_3} } }$
 * $\tilde \gamma_1 := \gamma_k {\restriction{ \closedint {\tilde t_3}{c_k} } }* \gamma_l {\restriction{ \closedint {c_{l-1} }{\tilde t_4} } }$

where $\gamma_k {\restriction{ \closedint {a}{b} } }$ denotes the restriction of $\gamma_k$ to $\closedint a b$, and $\gamma_1 * \gamma_2$ is the concatenation of $\gamma_1$ and $\gamma_2$.

From their definition, it follows that $\Img {\tilde \gamma_0 * \tilde \gamma_1} = \Img {\gamma_0}$.

From the definitions of Jordan curve and simple closed contour, it follows that $\tilde \gamma_0 * \tilde \gamma_1$ is a Jordan curve.

As the Jordan Curve Theorem says that the interior of a Jordan curve only depends on its image, it follows that $\Int {\tilde \gamma_0 * \tilde \gamma_1} = \Int {\gamma_0}$.

Let $\tilde \gamma_2: \closedint 0 1$ be the injective complex-differentiable function defined in the Lemma with $\map {\tilde \gamma_2} t \in \map {N_{r_1} }{ \map {\gamma_k}{c_k} }$ for all $t \in \openint 0 1$.

By the definitions of $\tilde t_3$ and $\tilde t_4$, it follows that $\map {\tilde \gamma_2} t \notin \Img {\tilde \gamma_0}$ for all $t \in \openint 0 1$.

By definition of Jordan curve, it follows that $\tilde \gamma_0 * \tilde \gamma_2$ is a Jordan curve.

By definition of contour, it follows that $\tilde \gamma_0 * \tilde \gamma_2$ is a parameterization of a simple closed contour.

By the definition of $\tilde t_1$, there exists some $t_0 \in \openint {c_{k-1} }{\tilde t_1}$ such that $\map {\gamma_k}{t_0} \notin \map {N_{r_0} }{ \map {\gamma_k}{c_k} }$.

By the definition of $\tilde t_2$, there exists some $t_1 \in \openint {\tilde t_2}{c_l}$ such that $\map {\gamma_l}{t_1} \notin \map {N_{r_0} }{ \map {\gamma_k}{c_k} }$.

By assumption, there exists $r \in \R_{>0}$ such that $\map v {\epsilon, t} \in \Int {\gamma}$ for all $\epsilon \in \openint 0 r$.

From Interior and Exterior of Partially Disjoint Jordan Curves, it follows that exists $\overline r \in \R_{>0}$ such that:


 * for all $z \in \map {N_{\overline r} }{\map {\gamma_k}{t_0} }$: $z \in \Int { \gamma }$, $z \in \Int { \tilde \gamma_0 * \tilde \gamma_2 }$
 * for all $z \in \map {N_{\overline r} }{\map {\gamma_l}{t_1} }$: $z \in \Int { \gamma }$, $z \in \Int { \tilde \gamma_0 * \tilde \gamma_2 }$

Set $\rho := \min \set {r, \overline r}$.

It follows that $\map v {\epsilon, t_0} \in \Int {\tilde \gamma_0 * \tilde \gamma_2}$ for all $\epsilon \in \openint 0 \rho$.

From Single Directed Smooth Curve Case, it follows that:


 * $\map {\gamma_0 * \gamma_2} {t_1} + \epsilon i S \map {\paren{ \tilde \gamma_0 * \tilde \gamma_2}'}{ t_1 } \in \Int { \tilde \gamma_0 * \tilde \gamma_2 }$ for all $\epsilon \in \openint 0 \rho$

From what we have shown above, it follows that:


 * $\map {\gamma} {t_1} + \epsilon i S \map {\gamma'}{ t_1 } \in \Int {\gamma}$ for all $\epsilon \in \openint 0 \rho$

which is what we wanted to prove.

Category: Orientation of Complex Contour]]