# Version of Jordan’s theorem for unbounded curves

Let $I$ denote the open interval $]0,1[$. Let $\gamma$ be a countinous map $I \to {\mathbb R}^2$. We say that $\gamma$ is stretched if it contains points that are arbitrarily close to the origin, and also points that are arbitrarily far from the origin.

Let $\gamma_1$ and $\gamma_2$ be two stretched maps whose images in the plane are disjoint, and such that $0\in \gamma_1(I)$. Does it follow that ${{\mathbb R}^2}\setminus (\gamma_1(I) \cup \gamma_2(I))$ is not path-connected ?

Update 10:35 To avoid the uninteresting counterexample explained in richard's comment below, I request that the images $\gamma_1(I)$ and $\gamma_2(I)$ be not only disjoint, but live in disjoint angular sectors with room between them.

Formally, I mean that there are four numbers $a,b,c,d$ with $0 \lt a \lt b \lt c \lt d \lt 2\pi$ such that $\gamma_1(I) \setminus \lbrace O \rbrace \subseteq S_{a,b}$ and $\gamma_2(I) \subseteq S_{c,d}$, where for any pair $x,y$ such that $0<x<y<2\pi$ we define the open angular sector $S_{x,y}$ as

$$S_{x,y}=\bigg\lbrace (r\cos(\theta),r\sin(\theta)) \bigg| r \gt 0, x < \theta < y \bigg\rbrace$$

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Aer $\gamma_1$ and $\gamma_2$ injective? –  23rd Dec 6 '12 at 9:01
I'm sorry that I misunderstand your question... –  lee Dec 6 '12 at 9:10
@richard : No, not in the uses I want to make of it, but a partial solution in this special case is welcome. –  Ewan Delanoy Dec 6 '12 at 9:10
Maybe you can compactify the R^2. –  lee Dec 6 '12 at 9:14
I think you need some additional assumption on $\gamma_1$ and $\gamma_2$(for example, injectivity) to exclude the following less interesting counter-example: $\gamma_1(I)$ is the $x$-axis and $\gamma_2(I)$ is the upper half plane. –  23rd Dec 6 '12 at 9:17
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I think the complement of the two paths is not path connected. What follows is an admittedly sketchy argument.

A continuous path $\beta :[0,1] \to R^2$ in the plane which does not pass through the origin takes on a minimum radius $r>0$ and a maximum radius $R>0$, since the radius $r(t)=|\beta(t)|$ is a continuous function from $[0,1]$ to the positive reals $(0,\infty)$.

Now the plane with the origin and sectors $S_{ab}$ and $S_{cd}$ removed, has two disjoint path components. If points $P,Q$ are chosen one in each component, then any path $\beta$ joining $P$ to $Q$ has to cross through one of the two sectors.

Say it crosses through sector $S_{ab}$. Then choose a point $X_1$ in $S_{ab}$ of radius less than $r$, and another point $X_2$ in $S_{ab}$ with radius greater than $R$, and such that each of $X_1$ and $X_2$ lies on the path $\gamma_1$, which we may do since $\gamma_1$ was assumed to be "strecthed" in sector $S_{ab}$. Then the part of $\gamma_1$ connecting $X_1$ to $X_2$ will necessarily cut through the part of the path $\beta$ as it crosses through sector $S_{ab}.$

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Thanks for this answer! I know how to fill the gap in your proof, I’ll post it as a separate answer when I have the time. –  Ewan Delanoy Dec 7 '12 at 8:21
Ewan: I think the main gap is in making precise that $\gamma$ "cuts through" one of the sectors. It should mean there is a subpath going from one bounding ray to the other, staying in the interior of the sector. I had tried via defining the map from $[0,1]$ into the unit circle via $\gamma/|\gamma|$, and then using that the circle is cut into two parts by the two sectors. I look forward to a gap-filled proof! –  coffeemath Dec 7 '12 at 9:23

As explained above, this answer is simply meant to fill the gaps in coffeemath’s answer.

Each and every angle has a unique measure in $[0,2\pi[$. This allows us to define a map ${\sf arg} : \Omega \to [0,2\pi[$ where $\Omega={\mathbb R}^2 \setminus \lbrace 0 \rbrace$.

The set $\Omega \setminus (S_{ab} \cup S_{cd})$ has two connected components :

$$C_1=\bigg\lbrace M \in \Omega \ \bigg| \ {\sf arg}(M) \in [d,2\pi[ \cup [0,a] \bigg\rbrace \ \text{and} \ C_2=\bigg\lbrace M \in \Omega\ \bigg| \ {\sf arg}(M) \in [b,c] \bigg\rbrace$$

Take $P\in C_1$ and $Q\in C_2$. Let $\beta : [0,1] \to {\mathbb R}^2$ joining $P$ and $Q$. We will show that $\beta$ must intersect $\gamma_1$ or $\gamma_2$, and hence that $P$ and $Q$ cannot lie in the same path connected component of ${{\mathbb R}^2} \setminus (\gamma_1(I)\cup \gamma_2(I))$. If $\beta$ passes through the origin, then it intersects $\gamma_1$ and we are done. So we henceforth assume that $\beta$ does not pass through $O$.

Step 1 : We can assume that ${\sf arg} ( P )=a,{\sf arg}(Q)=b$ and $\beta(]0,1[) \subseteq S_{ab}$.

Let $F_1=\lbrace t \in [0,1] | \beta(t) \in C_1 \rbrace$. Then $0\in F_1$ and $F_1$ is closed in $[0,1]$. Therefore $F_1$ contains $t_1={\sf sup}(F_1)$. If $\beta(t_1)$ were in the interior of $C_1$, then we would have $t_1+\varepsilon \in F_1$ for small enough $\varepsilon$, which is impossible. So $\beta(t_1)$ must lie on the boundary of $C_1$. Similarly, let $F_2=\lbrace t \in [t_1,1] | \beta(t) \in C_2 \rbrace$. Then $1\in F_2$ and $F_2$ is closed in $[0,1]$. Therefore $F_2$ contains $t_2={\sf inf}(F_2)$. If $\beta(t_2)$ were in the interior of $C_2$, then we would have $t_2-\varepsilon \in F_2$ for small enough $\varepsilon$, which is impossible. So $\beta(t_2)$ must lie on the boundary of $C_2$.

Using the affine transform $\beta’(t)=\beta(t_1+(t_2-t_1)t)$, we see that we can assume that $P$ lies on the boundary of $C_1$, $Q$ lies on the boundary of $C_2$, and $I=\beta(]0,1[) \subseteq S_{ab} \cup S_{cd}$. Since $I$ is connected, we must have $I \subseteq S_{ab}$ or $I \subseteq S_{cd}$. In the latter case we interchange $c,d$ with $a,b$, which finishes Step 1.

As in coffeemath’s answer, there are numbers $r,R$ with $0 \lt r \lt R$ and $\beta([0,1])$ is included in the annulus

$$A_{r,R}=\lbrace M \in \Omega | r \leq OM \leq R \rbrace$$

Since $\gamma_1$ is stretched, there are values $t_3$ and $t_4$ such that $X_1=\gamma_1(t_3)$ satisfies $OX_1 \leq r$ and $X_2=\gamma_1(t_4)$ satisfies $OX_2 \geq R$. We then have a continuous $\rho : [0,1] \to \Omega$ such that $\rho$ is a subpath of $\gamma$ (i.e. the image of $\rho$ is included in the image of $\gamma$), and $\rho(0)=X_1,\rho(1)=X_2$. It will suffice to show that $\beta$ intersects $\rho$.

Proceeding as in step 1, we see that

Step 2 : We can assume that $OX_1=r,OX_2=R$ and $\rho(]0,1[)$ is included in the interior of $A_{r,R}$.

These paths $\beta$ and $\rho$ take their values in the compact set $T_{a,b}=\lbrace M \in \Omega | r \leq OM \leq R, a \leq {\sf arg}(M) \leq b \rbrace$. Since we have a homeomorphism between $[0,1]^2$ and $T_{ab}$ (defined by $(x,y) \mapsto (r+(R-r)x).(\cos(a+(b-a)y),\sin(a+(b-a)y))$), it will suffice to show (one more time) the following (well-known) lemma :

“A point on every edge of a square” lemma. Let $ABCD$ be a square. Let $p_1$ be a path in that square joining a point of $[AB]$ to a point of $[CD]$, and $p_2$ be another path in that square joining a point of $[AC]$ to a point of $[BD]$. Then $p_1$ and $p_2$ intersect.

Proof of the lemma Suppose by contradiction that $p_1(s) \neq p_2(t)$ for any $(s,t)\in [0,1]^2$. Then, we can define a map

$$\Phi : [0,1]^2 \to S^1, (s,t) \mapsto \frac{p_1(s)-p_2(t)}{||p_1(s)-p_2(t)||}$$

By the théorème du relèvement (“pullback theorem” in English? unfortunately there seems to be no english version of the french wikipedia page), there is a continuous map $\phi : [0,1]^2 \to {\mathbb R}$ such that $\Phi(s,t) =(\cos(\phi(s,t)),\sin(\phi(s,t)))$ for any $(s,t)\in [0,1]^2$. There is an integer $k$ such that $\phi(0,0)\in [2k\pi-\frac{\pi}{2},2k\pi]$. Replacing $\phi$ with $\phi-2k\pi$, we can assume $k=0$.

Then $\phi(0,0) \in [-\frac{\pi}{2},0]$.

Walking from $p_2(0)$ to $p_2(1)$, we see that $\phi(0,1) \in [0,\frac{\pi}{2}]$.

Walking from $p_1(0)$ to $p_1(1)$, we see that $\phi(1,1) \in [\frac{\pi}{2},\pi]$.

Walking from $p_2(1)$ to $p_2(0)$, we see that $\phi(1,0) \in [\pi,\frac{3\pi}{2}]$.

Walking from $p_1(1)$ to $p_1(0)$, we see that $\phi(1,1) \in [\frac{3\pi}{2},2\pi]$, contradiction. qed

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It looks like it takes more work than I thought to "fill in the gaps". Looks good. +1 –  coffeemath Jan 21 '13 at 11:01