Jordan Curve Bounding Loop in Euclidean Plane

Theorem

Let $f : \closedint 0 1 \to \R^2$ be a loop in the Euclidean plane $\R^2$.

Let $\epsilon \in \R_{>0}$.

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

$\map d {\map \gamma t, \Img f} < \epsilon$

where $\map d {\map \gamma t, \Img f}$ denotes the Euclidean distance between $\map \gamma t$ and $\Img f$.

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

Proof

Cast a ray from an extreme point of $\Img f$

Let $\norm {\map f t} : \closedint 0 1 \to \R$ denote the Euclidean norm of $\map f t$, considered as a function of $t$.

From Norm on Vector Space is Continuous Function and Composite of Continuous Mappings between Metric Spaces is Continuous, it follows that $\norm {\map f t}$ is continuous .

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

Set $\ds K := \max_{t \mathop \in \closedint 0 1} \norm {\map f t}$.

From the Extreme Value Theorem it follows that $K$ is well-defined, and there exists $t_0 \in \closedint 0 1$ such that:

$\ds \norm {\map f {t_0} } = K$

By definition of closed ball, it follows that:

$\Img f \subseteq \map {B_K^-}{\bszero}$

where $\map {B_K^-}{\bszero}$ denotes an closed ball with center equal to the origin of $\R^2$.

Let $\LL : \hointr 0 \infty \to \R^2$ be a ray with start point $\map f {t_0}$ defined by:

$\map \LL r = r \map f {t_0} + \map f {t_0}$

Using Norm Axiom $\text N 2$: Positive Homogeneity, it follows that for all $r \in \openint 0 \infty$:

$\norm {\map \LL r} = \paren {1+r} \norm{ \map f {t_0} } > K$

which shows that $\map \LL r \notin \Img f$.

Also, the equality shows that $\norm {\map \LL r}$ is strictly increasing.

Find the intersection of the ray and a regular tesselation of hexagons

Set $s := \epsilon / 4$.

From Three Regular Tessellations:Hexagons, it follows that there exists a tessellation of $\R^2$ consisting of regular hexagons with side length $s$.

Let $S$ be the first side of the hexagons that $\LL$ intersects, going from $\map f {t_0}$ to infinity.

Suppose that $\LL$ intersects $S$ at a vertex, or $\map f {t_0}$ is a vertex.

In this case, translate the tessellation by a small quantity, say $\tuple {\dfrac s 6, \dfrac s 7}$, such that $\LL$ no longer first intersects the hexagons at a vertex.

Let $H , H'$ be the two regular hexagons which both have $S$ as a side, labelled such that $\map f {t_0}$ lies in the interior of $H$.

Suppose that $H'$ and $\Int f$ are not disjoint, which implies that $H'$ intersects the open ball $\map {B_r}{\bszero}$ with center equal to the origin $\bszero$ of $\R^2$.

In this case, let $S'$ be the second side of the hexagons that $\LL$ intersects, going from $\map f {t_0}$ to infinity.

The two regular hexagons, which have $S'$ as a side, are $H'$ and another hexagon which we label $H' '$.

As $\norm {\map \LL r}$ increases as $r$ increases, it follows that the distance between $H' '$ and $\Img f$ is larger than the distance between $H'$ and $\Img f$.

Hence, $H' '$ and $\Int f$ are disjoint.

Construct a curve consisting of sides of the hexagons

Set $S_0 := S'$ if $S'$ was defined, and if not, set $S_0 := S$.

It follows that of the two hexagons $H_0, H_0'$ which have $S_0$ as a side, the hexagon we label $H_0$ contains elements of $\Img f$, while $H_0'$ and $\Int f$ are disjoint.

Let $V_0 , V_1$ be the two vertices of the side $S_0$.

Let $\vec {V_0 V_1}$ be the vector quantity that represents the directed line segment with startpoint $V_0$ and endpoint $V_1$.

Label the two vertices such that $H_0$ lies to the right of $\vec {V_0 V_1}$, and $H_0'$ lies to the left of $\vec {V_0 V_1}$.

That is, if $\vec {V_0 V_1}$ is rotated in clockwise direction around $V_0$, then $V_1$ will first pass through $H_0$.

Next, we define a sequence of vertices $\sequence {V_n}_{n \in \N}$ in the tessellation by induction.

For $n \in \N_{\ge 2}$, let $H_{n-1}$ be the hexagon to the right of $\vec { V_{n-2} V_{n-1} }$ that contains elements of $\Img f$.

Let $H_{n-1}'$ be the hexagon to the left of $\vec { V_{n-2} V_{n-1} }$ that is disjoint with $\Img f$.

Let $H_n$ be the hexagon that has $V_{n-1}$ as a vertex, and does not have $\vec { V_{n-2} V_{n-1} }$ as a side.

If $H_n$ is disjoint with $\Img f$, define $V_n$ as the vertex shared by $H_n$ and $H_{n-1}$ that is not equal to $V_{n-1}$.

If $H_n$ is not disjoint with $\Img f$, define $V_n$ as the vertex shared by $H_n$ and $H_{n-1}'$ that is not equal to $V_{n-1}$.

In both cases, the hexagon to the right of $\vec { V_{n-1} V_{n} }$ will contain elements of $\Img f$, while the hexagon to the left of $\vec { V_{n-1} V_{n} }$ is disjoint with $\Img f$.

Show that the curve is a Jordan polygon

For all $\mathbf p \in \R^2$ that lies on $\vec { V_{n-1} V_{n} }$, it follows that $\mathbf p$ and some element of $\Img f$ lie in the hexagon to the right of $\vec { V_{n-1} V_{n} }$

From Distance between Points in Regular Hexagon, it follows that:

$\map d { \mathbf p, \Img f} \le 2 s = \epsilon / 2$

As $\Img f \subseteq \map {B_K}{\bszero}$, it follows by the triangle inequality that:

$\mathbf p \in \map {B_{K + 2s} }{\bszero}$

From Area of Circle, it follows that the area of $\map {B_{K + 2s} }{\bszero}$ is equal to $\pi \paren{ K + 2s }^2$.

From Area of Regular Hexagon, it follows that the area of each regular hexagon in the tessellation is equal to $\dfrac {3 \sqrt 3} 2 s^2$.

It follows that $\vec { V_{n-1} V_n }$ can be a side of at most $\dfrac { 2 \pi \paren{ K + 2s }^2 }{3 \sqrt 3 s^2}$ different hexagons.

Hence, there exists $n_0 , n_1 \in \N$ with $n_0 < n_1$ such that $V_{n_0} = V_{n_1}$.

Let $n_0$ be the smallest such number.

Suppose that $n_0 > 0$.

From the construction of $\vec { V_{n_0 - 1} V_{n_0} }$ and $\vec { V_{n_0} V_{n_0 + 1} }$, it follows that of the three hexagons $H_{n_0}, H_{n_0}', H_{n_0 + 1}$:

• either $H_{n_0}, H_{n_0 + 1}$ both contain elements of $\Img f$;
• or $H_{n_0}', H_{n_0 + 1}$ both are disjoint with $\Img f$.

As $\vec { V_{n_1 - 1} V_{n_1} }$ is not equal to $\vec { V_{n_0 - 1} V_{n_0} }$ or $\vec { V_{n_0} V_{n_0 + 1} }$ , it follows that the two hexagons that $\vec { V_{n_1 - 1} V_{n_1} }$ is a side of:

• either are equal to $H_{n_0}, H_{n_0 + 1}$, which both contain elements of $\Img f$;
• or are equal to $H_{n_0}', H_{n_0 + 1}$, which both are disjoint with $\Img f$.

In both cases, this contradicts the construction of $\vec { V_{n_1 - 1} V_{n_1} }$.

It follows that $n_0 = 0$.

Hence, $P = V_0 V_1 \cdots V_{n_1 - 2} V_{n_1 - 1}$ defines a polygon.

Prove the claims of the theorem

From Boundary of Polygon is Jordan Curve, it follows that there exists a Jordan curve $\gamma: \closedint 0 1 \to \R^2$ such that $\Img \gamma = \partial P$, where $\partial P$ denotes the boundary of $P$.

For all $t \in \closedint 0 1$, we have $\map \gamma t \in \partial P$, so from what we have shown above:

$\map d {\map \gamma t, \Img f} \le \epsilon / 2 < \epsilon$

From the Jordan Polygon Theorem, it follows that $\R^2 \setminus \partial P$ is a union of two disjoint open connected components $\Int \gamma$ and $\Ext \gamma$.

From Connected Open Subset of Euclidean Space is Path-Connected, it follows that $\Int \gamma$ and $\Ext \gamma$ are path components of $\R^2 \setminus \partial P$.

As $f$ is a loop that does not intersect $\partial P$, it follows that either $\Int f \subseteq \Int \gamma$, or $\Int f \subseteq \Ext \gamma$.

The ray $\LL$ with start point $\map f {t_0}$ crosses $\partial P$ at the side $V_0 V_1$.

As $\norm {\map \LL r}$ is strictly increasing, and:

$\ds \norm {\map \LL 0} = \norm {\map f {t_0} } = \max_{t \mathop \in \closedint 0 1} \norm {\map f t}$

it follows that $\LL$ does not cross $\partial P$ more than once. as $\map d {\partial P, \Img f} \le 2 s$.

From the Jordan Polygon Interior and Exterior Criterion, it follows that $\map f {t_0} \in \Int \gamma$.

As $\Int \gamma$ is path-connected, it follows that $\Img f \subseteq \Int \gamma$.

$\blacksquare$