Single Point Characterization of Simple Closed Contour/Lemma 1

Lemma

Let $C$ be a simple closed contour in the complex plane $\C$ with parameterization $\gamma: \closedint a b \to \C$.

Let $S \in \set {-1, 1}$.

Let $\set {c_0, \ldots, c_N }$ be a subdivision of $\closedint a b$ with $N \in \N$ such that $\gamma$ is complex-differentiable at all $t \in \openint {c_{n - 1} } {c_n}$ for $n \in \set {1, \ldots, N}$.

For all $\epsilon \in \R_{>0}$ and $t \in \openint a b$ for which $\gamma$ is complex-differentiable at $t$, define:

$\map v {t, \epsilon} := \map \gamma t + \epsilon i S \map {\gamma '} t$

Let $k \in \set {1, \ldots, N}$ and $t_0 \in \openint {c_{k - 1} } {c_k}$ such that there exists $r \in \R_{>0}$, where for all $\epsilon \in \openint 0 r$:

$\map v {t_0, \epsilon} \in \Int C$

Then for all $t_1 \in \openint {c_{k - 1} } {c_k}$, there exists $r \in \R_{>0}$ such that for all $\epsilon \in \openint 0 r$:

$\map v {t_1, \epsilon} \in \Int C$

Proof

Let $\Img C$ denote the image of $C$.

From Normal Vectors Form Space around Simple Complex Contour, it follows that there exists $r_1 \in \R_{>0}$ such that for all $\epsilon \in \openint 0 {r_1}$, we have $\map v {t_1, \epsilon} \notin \Img C$.

Complex Plane is Homeomorphic to Real Plane shows that we can identify the complex plane $\C$ with the real plane $\R^2$ by the homeomorphism $\map \phi {x, y} = x + i y$.

Interior of Simple Closed Contour is Well-Defined shows that $\Img C$ can be identified with the image of a Jordan curve $g: \R^2 \to \R^2$.

From the same theorem, it follows that $\Int C$ can be identified with the interior of $g$.

From the Jordan Curve Theorem, it follows that $\Int C$ is an open connected component of $\C \setminus \Img C$.

From Connected Open Subset of Euclidean Space is Path-Connected, it follows that $\Int C$ is a path component of $\C \setminus \Img C$.

For all $\epsilon_0, \epsilon_1 \in \openint 0 {r_1}$ with $\epsilon_0 < \epsilon_1$, there is a path $\sigma : \closedint {\epsilon_0} {\epsilon_1} \to \C \setminus \Img C$ between $\map v {t_1, \epsilon_0}$ and $\map v {t_1, \epsilon_1}$, defined by the line segment:

for all $s \in \closedint {\epsilon_0} {\epsilon_1} : \map \sigma s = \map v {t_1 , s}$

If $\map v {t_1, \epsilon_1} \in \Int C$ for one value of $\epsilon_1 \in \openint 0 {r_1}$, it follows that $\map v {t_1, \epsilon} \in \Int C$ for all $\epsilon \in \openint 0 {r_1}$, which is what we wanted to prove.

Aiming for a contradiction, suppose instead that $\map v {t_1, \epsilon} \notin \Int C$ for all $\epsilon \in \openint 0 {r_1}$.

Let $\mathbb I$ be the closed real interval with endpoints $t_0$ and $t_1$.

Set:

$t_2 := \sup \set {t \in \mathbb I : \exists r_2 \in \R_{>0} \forall \epsilon \in \openint 0 {r_2} : \map v {t_2, \epsilon} \in \Int C}$

From Normal Vectors Form Space around Simple Complex Contour, there exists $\tilde r_2 \in \R_{>0}$ such that for all $\epsilon \in \openint 0 {\tilde r_2}$, we have $\map v {t_2, \epsilon} \notin \Img C$.

Set $\epsilon_2 := \dfrac {\tilde r_2} 2$.

From Continuous Image of Compact Space is Compact, it follows that $\Img C$ is compact in $\C$.

From Singleton in Normed Vector Space is Closed, it follows that $\set {\map v {t_2, \epsilon_2} }$ is closed.

From Distance between Disjoint Compact Set and Closed Set in Metric Space is Positive, it follows that there exists $h \in \R_{>0}$ such that the open disk $\map {B_h} {\map v {t_2, \epsilon_2} }$ is disjoint with $\Img C$.

From Open Ball is Simply Connected, it follows that $\map {B_h} {\map v {t_2, \epsilon_2} }$ is path-connected.

As $\Int C$ is a path component, it follows that either $\map {B_h} {\map v {t_2, \epsilon_2} } \subseteq \Int C$, or $\map {B_h} {\map v {t_2, \epsilon_2} } \cap \Int C = \O$.

As $\gamma'$ is continuous by definition of parameterization, it follows that:

$\ds \lim_{t \mathop \to t_2} \map v {t, \epsilon_2} = \map v {t_2, \epsilon_2}$

From the definition of $t_2$ as a supremum, for all $n \in \N$ there exist:

$\tilde t_0 \in \closedint {t_2 - \dfrac 1 n} {t_2}$, $\tilde t_1 \in \closedint {t_2} {t_2 + \dfrac 1 n}$

such that $\map v {\tilde t_0 , \epsilon_2} \in \Int C$, and $\map v {\tilde t_1, \epsilon_2} \notin \Int C$.

Chose $n$ sufficiently large, so $\map v {\tilde t_0, \epsilon_2}, \map v {\tilde t_1, \epsilon_2} \in \map {B_h} {\map v {t_2, \epsilon_2} }$.

Given that $\map {B_h} {\map v {t_2, \epsilon_2} }$ is either a subset of or disjoint with $\Int C$, this is a contradiction.

$\blacksquare$