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:


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

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 :


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

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

As $r_1 < r_0$, this implies that:

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

Category: Orientation of Complex Contour]]