Component of Complement of Jordan Curve has Curve as Boundary

== 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:
* Closed Real Interval is Compact
* Continuous Image of Compact Space is Compact
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 , $W$ 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 ===
 $\partial U \subsetneq J$.
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 $o$.
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:
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:
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:
:$\partial U = J$
== Sources ==
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Category:Jordan Curves