Component of Complement of Jordan Curve has Curve as Boundary

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Theorem

Let $\phi : \closedint 0 1 \to \R^2$ be a Jordan curve.

Let $J = \phi \closedint 0 1$ be the image of $\phi$.

Suppose that $\R^2 \setminus J$ has at least two distinct components.


Then, for any component $U$ of $\R^2 \setminus J$:

$\partial U = J$

where $\partial U$ denotes the boundary of $U$.


Proof

By:

it follows that $J$ is compact.

Therefore, by Compact Subspace of Hausdorff Space is Closed:

$J$ is closed in $\R^2$

so by definition of closed:

$\R^2 \setminus J$ is open in $\R^2$

By definition of locally connected:

Each component of $\R^2 \setminus J$ is open


Let $W$ be the union of all components of $\R^2 \setminus J$, excluding $U$.

By Open Set Axiom $\paren {\text O 1 }$: Union of Open Sets, $W$ is open.

Additionally:

$U \cup W = \R^2 \setminus J$
$U \cap W = \O$

By Open Set Disjoint from Set is Disjoint from Closure and the definition of closure:

$\paren {U \cup \partial U} \cap W = \O$

By Intersection Distributes over Union, Set is Subset of Union, Subset of Empty Set:

$\partial U \cap W = \O$


Combining the above with Set is Open iff Disjoint from Boundary, we get:

$\partial U \cap \paren {\R^2 \setminus J} = \O$

That is:

$\partial U \subseteq J$


Jordan Arc

Aiming for a contradiction, suppose $\partial U \subsetneq J$.

That is, there is some $\bsx_0 \in J \setminus U$.

As $\bsx_0 \notin U^-$ and Set is Closed iff Equals Topological Closure:

$\bsx_0 \in \paren {U^-}^e$

where $U^e$ denotes the exterior of $U$.

But then, by definition of exterior:

$\bsx_0$ is an interior point of $\R^2 \setminus U^-$

That is:

There is an open neighborhood $N_{\bsx_0}$ such that $N_{\bsx_0} \subseteq \R^2 \setminus U^-$

By definition of continuous:

$\phi^{-1} \sqbrk {N_{\bsx_0}}$ is a neighborhood of all $t_0 \in \closedint 0 1$ such that $\map \phi {t_0} = \bsx_0$

If $t_0 \in \openint 0 1$, let $\epsilon > 0$ be such that:

$\openint {t_0 - \epsilon} {t_0 + \epsilon} \subseteq \phi^{-1} \sqbrk {N_{\bsx_0}}$

where $a < t_0 - \epsilon < t_0 + \epsilon < b$, which exists by definition of neighborhood and of topology induced by metric.

Then, $\phi^* : \closedint 0 1 \to J$ defined as:

$\map {\phi^*} t = \begin{cases} \map \phi {2 t + \paren {1 - 2 t} \paren {t_0 + \epsilon}} & : t \in \closedint 0 {\dfrac 1 2} \\ \map \phi {\paren {2 t - 1} \paren {t_0 - \epsilon}} & : t \in \closedint {\dfrac 1 2} 1 \end{cases}$

is a continuous injection, and thus a Jordan arc.

Additionally:

$\phi^* \closedint 0 1 \supseteq J \setminus N_{\bsx_0} \supseteq \partial U$


Now, suppose $t_0 = 0, 1$.

By definition of Jordan curve, it immediately follows that both $0, 1$ satisfy $\map \phi {t_0} = \bsx_0$.

Therefore, there is some $\epsilon > 0$ such that:

$\hointr 0 \epsilon \cup \hointl {1 - \epsilon} 1 \subseteq \phi^{-1} \sqbrk {N_{\bsx_0}}$

Then, $\phi^* : \closedint 0 1 \to J$ defined as:

$\map {\phi^*} t = \map \phi {\paren {1 - t} \epsilon + t \paren {1 - \epsilon}}$

is also a Jordan arc, and satisfies:

$\phi^* \closedint 0 1 \supseteq \partial U$


Extension

By Compact Subspace of Metric Space is Bounded:

$J$ is bounded

Therefore, by Complement of Bounded Set has Exactly One Unbounded Component, and the assumption that there are at least two components of $\R^2 \setminus J$:

There exists a bounded component of $\R^2 \setminus J$

If $U$ is bounded:

Let $o \in U$ be arbitrary

Otherwise:

Let $o \in \R^2 \setminus J$ be contained in a bounded component.


By definition of bounded, it follows that there exists some $K > 0$ such that:

$J \subseteq \map {B_K} o$

where $\map {B_K} o$ denotes the open ball with radius $K$ and center $o$.


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From the proof of Complement of Bounded Set has Exactly One Unbounded Component, it follows that $\partial \map {B_K} o$ is contained in the unique unbounded component of $\R^2 \setminus J$.

As $\phi^*$ is a injection, by Injection to Image is Bijection, we can consider:

$\phi^* : \closedint 0 1 \to A$

as a bijection, where $A = \phi^* \closedint 0 1$.

Then, ${\phi^*}^{-1}$ is a mapping by definition of bijection.


Additionally, since $\closedint 0 1$ is compact by Closed Real Interval is Compact, by Continuous Bijection from Compact to Hausdorff is Homeomorphism:

${\phi^*}^{-1}$ is continuous

Therefore, by the Tietze Extension Theorem, there exists a continuous extension of ${\phi^*}^{-1}$:

$r : D \to \R$

where $D = {\map {B_K} o}^-$.

By Minimum Rule for Continuous Functions and Maximum Rule for Continuous Functions, $r^* : D \to \closedint 0 1$, defined as:

$\map {r^*} \bsx = \begin{cases} 0 & : \map r \bsx < 0 \\ \map r \bsx & : 0 \le \map r \bsx \le 1 \\ 1 & : \map r \bsx > 1 \end{cases}$

is also continuous.


Fixed Point

We will now define $q : D \to D \setminus \set o$.

If $U$ is bounded, then:

$\map q \bsz = \begin{cases} \map {\phi^*} {\map {r^*} \bsz} & : \bsz \in U^- \\ \bsz & : \bsz \in D \setminus U \end{cases}$

If $U$ is unbounded, then:

$\map q \bsz = \begin{cases} \bsz & : \bsz \in U^- \\ \map {\phi^*} {\map {r^*} \bsz} & : \bsz \in D \setminus U \end{cases}$

In order for these to be well-defined, it is required that:

$\map {\phi^*} {\map {r^*} \bsz} = \bsz$

for all $\bsz \in \partial U$.


We have already shown that $\partial U \subseteq A$, so:

\(\ds \map {\phi^*} {\map {r^*} \bsz}\) \(=\) \(\ds \map {\phi^*} {\map { {\phi^*}^{-1} } \bsz}\) As $r$ extends $\phi^*$, and is thus on $\closedint 0 1$
\(\ds \) \(=\) \(\ds \bsz\)


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Additionally, $q$ is continuous.

Finally, to show that $o \notin \Img q$, observe that $o$ is always contained in the set such that:

$\map q o = \map {r^*} o \in A$


Now, we define $p : D \setminus \set o \to \partial D$ as:

$\map p \bsx = o - \dfrac K {\norm {\bsx - o}} \paren {\bsx - o}$

where $K$ is the radius of $D$.

$p$ is clearly continuous by Combination Theorem for Continuous Mappings on Metric Space.


We want to show that $p \circ q : D \to D$ has no fixed point.

As $\Img p \subseteq \partial D$, it is clear that any fixed point must be in $\partial D$.

As such, let $\bsx \in \partial D$ be arbitrary.

As $\bsx$ is in the unbounded component of $\R^2 \setminus J$, it is either:

In $D \setminus U$ if $U$ is bounded
In $U^-$ if $U$ is unbounded

Thus, in either case:

$\map q \bsx = \bsx$

Therefore:

\(\ds \map {\paren {p \circ q} } \bsx\) \(=\) \(\ds \map p \bsx\)
\(\ds \) \(=\) \(\ds o - \dfrac K {\norm {\bsx - o} } \paren {\bsx - o}\) Definition of $p$
\(\ds \) \(=\) \(\ds o - \paren {\bsx - o}\) $\bsx \in \partial D$
\(\ds \) \(=\) \(\ds 2 o - \bsx\)
\(\ds \leadsto \ \ \) \(\ds \norm {\map {\paren {p \circ q} } \bsx - \bsx}\) \(=\) \(\ds \norm {2 o - 2 \bsx}\)
\(\ds \) \(=\) \(\ds 2 K\) $\bsx \in \partial D$
\(\ds \leadsto \ \ \) \(\ds \map {p \circ q} \bsx\) \(\ne\) \(\ds \bsx\) Norm Axiom $\text N 1$: Positive Definiteness

Therefore, $p \circ q : D \to D$ has no fixed point.

But by Brouwer's Fixed Point Theorem, it must have a fixed point.

This is a contradiction, so our assumption that $\partial U \subsetneq J$ was false.

Thus, by Proof by Contradiction:

$\partial U = J$

$\blacksquare$


Sources

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