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Preliminaries

A Jordan map is a continuous map $f: [0,1] \rightarrow \mathbb{R}^2$ such that

  1. $f(0) = f(1)$
  2. the restriction of $f\ $ to $[0,1)$ is injective

A Jordan curve is a subset $\gamma$ of $\mathbb{R}^2$ such that there is a Jordan map $f$ with $\gamma = f(\mathbb{R})$.

Jordan curve theorem: For each Jordan curve $\gamma$ there are subsets $A,B$ of $\mathbb{R}^2$ such that

  1. $\mathbb{R}^2 = A \cup B \cup \gamma$
  2. $A$, $B$, and $\gamma$ are disjoint.
  3. Any continuous path connecting a point in $A$ to a point in $B$ intersects $\gamma$.
  4. $\gamma$ is the boundary of both $A,B$.
  5. One of $A,B$ is bounded and the other is unbounded.

How is this carried over to curves that bisect $\mathbb{R}^2$ into two unbounded components $A, B$?

Especially:

What would be the definition of a bisecting map? Is it – for example – just an injective continuous map $f:\mathbb{R}\rightarrow \mathbb{R}^2$ such that $\lim\limits_{|x|\to \infty} |f(x)| \rightarrow \infty$?

Such a definition given it would be straight forward to define a bisecting curve and to state a bisecting curve lemma: For each bisecting curve $\gamma$ there are subsets $A,B$ of $\mathbb{R}^2$ such that

  1. $\mathbb{R}^2 = A \cup B \cup \gamma$
  2. $A$, $B$, and $\gamma$ are disjoint.
  3. Any continuous path connecting a point in $A$ to a point in $B$ intersects $\gamma$.
  4. $\gamma$ is the boundary of both $A,B$.
  5. Both $A,B$ are unbounded.

Would the proof of such a bisecting curve lemma follow essentially the tracks of the Jordan curve theorem?

Or was it – historically – the other way around: first an (easier) bisecting curve lemma was proved, and only after that the (more complicated) Jordan curve theorem? (I have to admit that this is not clear to me.)

MAIN QUESTION

Consider homeomorphisms $F: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ such that $F(\mathbb{R}\times\lbrace 0 \rbrace)$ is a bisecting curve. (Probably this is the case for all homeomorphisms?) Let $F \sim G$ iff $F(\mathbb{R}\times\lbrace 0 \rbrace) = G(\mathbb{R}\times\lbrace 0 \rbrace)$.

Is there a "naturally" definable partial order on $[F]$ ("is simpler than") such that there is a unique minimal (i.e. simplest) $F_0$ in $[F]$?

This $F_0$ would be the unique simplest map that distorts the plane taking $\mathbb{R}\times\lbrace 0 \rbrace$ to $F_0(\mathbb{R}\times\lbrace 0 \rbrace)$.

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The answer is yes to all your main questions, and the generalizations follow directly from the Jordan curve theorem. –  George Lowther Sep 22 '11 at 0:00
    
So please tell me: what is the partial order I ask for in my third main question. –  Hans Stricker Sep 22 '11 at 0:02
    
The Euclidean plane is homeomorphic to the sphere with one point removed, by stereographic projection. So, the Jordan curve theorem is equivalent to saying that a Jordan curve on the sphere divides it into two components. There is no "infinite" component, until you remove a point and map the sphere to a plane, in which case the infinite component is the one containing the point at infinity (the point removed). –  George Lowther Sep 22 '11 at 0:03
1  
Your third question is a bit different, but the answer is still yes, almost. The simplest map will be the one given by the Riemann mapping theorem applied to each of the two connected components. It's unique up to a one parameter family of maps (horizontal translations, in the way you have specified things here). –  George Lowther Sep 22 '11 at 0:09
1  
@George: Couldn't you give this as an answer? –  Hans Stricker Sep 22 '11 at 0:14

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