User:Anghel/Sandbox

Proof
This proof assumes that the derivative $f' : U \to \C$ is continuous.

This assumption is always true, as can be seen from Holomorphic Function is Continuously Differentiable.

However, the proof of Holomorphic Function is Continuously Differentiable depends on Cauchy's Integral Formula, which again depends on the Cauchy-Goursat Theorem.

It follows that this proof is circular, unless $f$ is known to be continuously differentiable.

Suppose that $C$ is a simple closed staircase contour.

Corollary
Let $f: U \to \C$ be a holomorphic function, where $U \subseteq \C$ is an open set.

Let $C$ be a simple closed contour in $U$.

Let $\Int C \subseteq U$, where $\Int C$ denotes the interior of $C$.

Then:


 * $\ds \oint_C \map f z \rd z = 0$

Theorem
Let $C$ be a simple closed contour in $U$, where $U \subseteq \C$ is an open set.

Let $\Int C \subseteq U$, where $\Int C$ denotes the interior of $C$.

Then there exists a simply connected domain $V$ such that $\Int C \subseteq V \subseteq U$, and $C$ is a contour in $V$.

Proof
By Complex Plane is Homeomorphic to Real Plane, the function $\phi : \R^2 \to \C$, defined by $\map \phi {x, y} = x + i y$, is a homeomorphism between $\R^2$ and $\C$.

By Interior of Simple Closed Contour is Well-Defined, there exists a Jordan curve $f : \closedint 0 1 \to \R^2$ with $\Img C = \phi \sqbrk {\Img f}$, and $\Int C = \phi \sqbrk {\Int f}$.

As $\phi$ is bijective, it follows that $\Img f = \phi^{-1} \sqbrk {\Img C}$, and $\Int f = \phi^{-1} \sqbrk {\Int C}$.

Let $\mathbb S^1$ denote the unit circle in $\R^2$ whose center is at the origin $\mathbf 0$ of $\R^2$.

Let $\map {B_1} { \mathbf 0 }$ denote the open ball in $\R^2$ with radius $1$ and center $\mathbf 0$, and let $\map {B_1^-} { \mathbf 0 }$ denote the closed ball in $\R^2$ with radius $1$ and center $\mathbf 0$.

By the Jordan-Schönflies Theorem, there exists a homeomorphism $\psi: \R^2 \to \R^2$ such that $\psi \sqbrk {\Img f} = \mathbb S_1$, and $\psi \sqbrk{ \Int f } = \map{B_1}{\mathbf 0}$.

By Composite of Homeomorphisms between Metric Spaces is Homeomorphism, $\psi \circ \phi^{-1} : \C \to \R^2$ is a homeomorphism.

It follows that $\mathbb S_1 = \psi \circ \phi^{-1} \sqbrk {\Img C}$, and $\map{B_1}{\mathbf 0} = \psi \circ \phi^{-1} \sqbrk {\Int C}$.

By definition of closed ball, it follows that $\map {B_1^-}{ \mathbf 0} = \mathbb S_1 \cup \map{B_1}{\mathbf 0}$.

By Closed Ball in Euclidean Space is Compact, $\map {B_1^-}{\mathbf 0}$ is compact.

By Complement of Open Set in Complex Plane is Closed, $\relcomp \C U$ is closed in $\C$.

Set $K = \psi \circ \phi^{-1} \sqbrk { \relcomp \C U }$.

By Continuity of Mapping between Metric Spaces by Closed Sets, $K$ is closed in $\R^2$.

As $\psi \circ \phi^{-1}$ is bijective, and $\Img C \cup \Int C$ and $\relcomp \C U$ are disjoint in $\C$, it follows that $\map{B_1^-}{\mathbf 0}$ and $K$ are disjoint in $\R^2$.

By Distance between Disjoint Compact Set and Closed Set in Metric Space is Positive, the distance between $\map{B_1^-}{\mathbf 0}$ and $K$ is equal to some $\epsilon \in \R_{>0}$.

For each $\mathbf x \in \map { B_{1+\epsilon} }{\mathbf 0}$, the distance between $\mathbf x$ and $\map {B_1^-}{\mathbf 0}$ is smaller than $\epsilon$.

By definition of distance, it follows that $\mathbf x \notin K$, so $\map { B_{1+\epsilon} }{\mathbf 0}$ and $K$ are disjoint in $\R^2$.

By Open Ball is Open Set in Normed Vector Space, $\map { B_{1+\epsilon} }{\mathbf 0}$ is open in $\R^2$.

By Open Ball is Simply Connected, $\map { B_{1+\epsilon} }{\mathbf 0}$ is simply connected.

Set $V = \phi \circ \psi^{-1} \sqbrk {\map { B_{1+\epsilon} }{\mathbf 0} } \subseteq \C$.

By Simple Connectedness is Preserved under Homeomorphism, it follows that $V$ is simply connected.

By definition of continuity, $V$ is open in $\C$.

By definition of simply connected domain, $V$ is a simply connected domain.

As $\map {B_1^-}{\mathbf 0} \subseteq \map { B_{1+\epsilon} }{\mathbf 0}$, and $\phi \circ \psi^{-1}$ is bijective, it follows that $\Img C \cup \Int C \subseteq V$.

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