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 $t \in \openint{\tilde t_3}{\tilde t_4}$

Category: Orientation of Complex Contour]]