Jordan Curve Characterization of Simply Connected Set

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Theorem

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


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

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


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


Proof

Sufficient condition

Suppose that $D$ is simply connected.

Let $\gamma : \closedint 0 1 \to \R^2$ be a Jordan curve with $\Img \gamma \subseteq D$.

By definition of simple connectedness, there exists a constant loop $c : \closedint 0 1 \to D$ such that $\gamma$ and $c$ are path-homotopic in $D$.

Let $H : \closedint 0 1 \times \closedint 0 1 \to D$ be the path homotopy between $\gamma$ and $c$.

From Interior of Jordan Curve is Subset of Image of Null-Homotopy, it follows that:

$\Int \gamma \subseteq \Img H$

As $\Img H \subseteq D$, it follows that:

$\Int \gamma \subseteq D$

$\Box$


Necessary condition

Suppose that for all Jordan curves $\gamma : \closedint 0 1 \to \R^2$ with $\Img \gamma \subseteq D$, we have $\Int \gamma \subseteq D$.

Let $f : \closedint 0 1 \to D$ be a loop in $D$.

By definition of closed set, it follows that the complement $\relcomp {\R^2} D$ is closed in $\R^2$.

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

From Continuous Mappings preserve Compact Subsets, it follows that $\Img f$ is compact.

From Distance between Disjoint Compact Set and Closed Set in Metric Space is Positive, it follows that:

$\map d {\Img f, \relcomp {\R^2} D} > 0$

where :$\map d {X,Y}$ denotes the Euclidean distance between the sets $X$ and $Y$.

From Jordan Curve Bounding Loop in Euclidean Plane, there exists a Jordan curve $\gamma : \closedint 0 1 \to \R^2$ such that $\Img f \subseteq \Int \gamma$, and for all $t \in \closedint 0 1$:

$\map d{\map \gamma t, \Img f} < \map d {\Img f, \relcomp {\R^2} D} / 2$

It follows that $\Img \gamma \subseteq D$.

From Interior of Jordan Curve is Simply Connected, it follows that $\Img \gamma$ is simply connected.

By definition of simple connectedness, there exists a constant loop $c : \closedint 0 1 \to D$ such that $f$ and $c$ are path-homotopic in $\Int \gamma$.

Let $H : \closedint 0 1 \times \closedint 0 1 \to \Int \gamma$ be the path homotopy between $f$ and $c$ in $\Int \gamma$.

As $\Img \gamma \subseteq D$, it follows that the $H$ is also a path homotopy between $f$ and $c$ in $D$.

By definition of simple connectedness, it follows that $D$ is simply connected.

$\blacksquare$


Sources

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