User:Anghel/Sandbox

Theorem
Let $D \subseteq \R^2$ be an open path-connected subset of the Euclidean plane.

Then $D$ is simply connected, the following condition holds:


 * For all Jordan curves $f : \closedint 0 1 \to \R^2$ with $\Img f \subseteq D$, we have $\Int f \subseteq D$.

Here $\Img f$ denotes the image of $f$, and $\Int f$ denotes the interior of $f$.

Sufficient condition
Suppose $D$ is simply connected.

Let $f : \closedint 0 1 \to D$ be a Jordan curve.

By definition of loop, $f$ is a loop in $D$.

By definition of simple connectedness, there exists a path homotopy $H_0 : \closedint 0 1 \times \closedint 0 1 \to D$ between $f$ and a constant loop $c_0: \closedint 0 1 \to \set { \map f 0 }$.

It follows that $H_0$ is a null-homotopy between $f$ and $c_0$.

Let $\sim$ be the equivalence relation on $\closedint 0 1$ defined by:

From Simple Loop in Hausdorff Space is Homeomorphic to Quotient Space of Interval, it follows that $\Img f$ is homeomorphic to $\closedint 0 1 / \sim$.

Let the homeomorphism between $\closedint 0 1 / \sim$ and $\Img f$ be $h: \closedint 0 1 / \sim \to \Img f$.

It follows that $h$ is a continuous injective mapping, defined by:


 * $\map h {\eqclass s \sim } = \map f s$

where $\eqclass x \sim$ denotes the equivalence class defined by $\sim$.

Define $H : \paren{ \closedint 0 1 / \sim } \times \closedint 0 1 \to D$ by:


 * $\map H { \eqclass s \sim,t } = \map {H_0}{s,t}$

which is well-defined, as $\map {H_0}{0,t} = \map {H_0}{1,t}$ for all $t \in \closedint 0 1$.

From Composite of Continuous Mappings is Continuous, it follows that $H$ is continuous.

Hence, $H$ is a null-homotopy between $h$ and a constant loop $c: \closedint 0 1 / \sim \to \set { \map f 0 }$.

From Closed Real Interval is Compact, it follows that $\closedint 0 1$ is compact.

From Continuous Image of Compact Space is Compact, it follows that $\Img f$, as well as $\closedint 0 1 / \sim$, are compact.

From Topological Product of Compact Spaces, it follows that $\paren{ \closedint 0 1 / \sim } \times \closedint 0 1$ is compact.

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

From Compact Subspace of Metric Space is Bounded, it follows that $\Img H$ is bounded.

From the Jordan Curve Theorem, it follows that $\R^2 \setminus \Img f$ is a union of two disjoint connected components.

These components are the interior $\Int f$, which is bounded, and the exterior $\Ext f$, which is unbounded.

By definitions of bounded and unbounded, it follows that there exists some $\mathbf a \in \Ext f \setminus \Img H$.

Let $\mathbf b \in \R^2 \setminus \Img H$.

From Borsuk Null-Homotopy Lemma:Corollary, it follows that $\mathbf a$ and $\mathbf b$ lie in the same component of $\R^2 \setminus \Img f$.

That is, $\mathbf b \in \Ext f$.

It follows that if $\mathbf x \in \Int f$, then $\mathbf x \in \Img H$.

As $\Img H \subseteq D$, we conclude that $\Int f \subseteq D$.

qed}}