User:Anghel/Sandbox

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 $V \subseteq U$, 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 Blah, there exists a Jordan curve $f : \closedint 0 1 \to \R^2$ with $\Int f = \phi^{-1} \sqbrk {\Int C}$.

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

By Normed Vector Space is Finite Dimensional iff Unit Sphere is Compact, $\mathbf S_1$ is compact, as the Euclidean space $\R^2$ has dimension $2$.