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

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

Set:


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

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