18

A Jordan curve is a continuous closed curve in $\Bbb R^2$ which is simple, i.e. has no self-intersections. The Jordan curve theorem states that the complement of any Jordan curve has two connected components, an interior and an exterior.

Now let's define an unbounded curve to be a continuous map $f: (-\infty,\infty)\to\Bbb R^2$ such that $f((-\infty,0))$ and $f((0,\infty))$ are both unbounded. My question is, does the complement of a simple unbounded curve always have two connected components? It seems intuitively true, since you'd expect the curve to have two sides, but considering how long it took to prove the Jordan curve theorem, things may not be as straightforward as they appear.

Any help would be greatly appreciated.

Thank You in Advance.

EDIT: As @dfeuer suggested, let's also require that the curve goes off to infinity in both directions. To make this precise, let's say that there exists two lines $L_1$ and $L_2$, parametrized by $L_1(t) = (a_1 + b_1 t, c_1 + d_1 t)$ and $L_2(t) = (a_2 + b_2 t, c_2 + d_2 t)$, such that the limit of $d(f(t), L_1(t))$ as $t$ goes to $-\infty$ is $0$, and the limit of $d(f(t), L_2(t))$ as $t$ goes to $\infty$ is $0$. Under that condition, does the complement of the curve have two connected components?

Stefan Hamcke
  • 27,733
Keshav Srinivasan
  • 10,204
  • 2
    Your title might be a little misleading, because I would call any curve whose end-points don't join up (like a letter C for example) a non-closed curve - these could maybe be called unbounded curves? I think the answer is going to be yes because you can compactify $\mathbb{R}^2$ to get a $2$-sphere, and the Jordan Curve Theorem holds on the $2$-sphere, but maybe somebody can make this precise. – mdp Aug 09 '13 at 15:20
  • 1
    @StefanH.: Nice example. And for four components, try $t \mapsto (\tan^{-1} t, t^2 \sin t^2)$. I have a vague picture in my head as to how we could get infinitely many components, but not yet a precise example. – Nate Eldredge Aug 09 '13 at 15:56
  • 1
    @MattPressland: In light of the counterexamples above, I guess the issue is that $(-\infty, \infty)$ embeds in its compactification $S^1$, and $\mathbb{R}^2$ embeds in its compactification $S^2$, but $f$ may not extend continuously to a map $S^1 \to S^2$, so we cannot apply the Jordan Curve Theorem. – Nate Eldredge Aug 09 '13 at 16:03
  • @NateEldredge I can believe that - I did wonder if my approach might be a little naive. – mdp Aug 09 '13 at 16:06
  • @NateEldredge: Thanks, but yours is one component nicer :-) – Stefan Hamcke Aug 09 '13 at 16:09
  • 1
    A modification that I think does the trick is to require the curve to approach infinity in both directions. That should fix the compactification argument, I believe. – dfeuer Aug 09 '13 at 16:22
  • @keshav Srinivasan From where did you get this problem? Any textbook? Please inform. – Supriyo Aug 09 '13 at 16:42
  • 4
    @dfeuer: I agree with that. I think the precise condition for a map $f : X \to Y$ to extend ctsly to a map $\tilde f : \tilde X \to \tilde Y$ of the 1-pt compactifications by $\tilde f( \infty_X) = \infty_Y$ is precisely that $f$ be proper -- which in this case amounts to requiring that the norm of $f$ approach infinity in both directions. – Mike F Aug 09 '13 at 17:01
  • @Samprity I didn't get it from anywhere. Just idle curiosity, based on another question I had about the boundaries of simply connected open sets. I may post that question as well on this site. – Keshav Srinivasan Aug 09 '13 at 17:07
  • @Mike Do you have any idea how to prove that? – Keshav Srinivasan Aug 09 '13 at 17:07
  • Also, does anyone know where I can find a proof that the Jordan curve theorem holds on the 2-sphere? – Keshav Srinivasan Aug 09 '13 at 17:09
  • 1
    @Keshav: It's essentially by definition. For the forward implication, suppose that $f : X \to Y$ is proper, $X,Y$ compact Hausdorff. This says $f^{-1}(K)$ is compact for $K \subset Y$ compact. Make the extension $\tilde f$. There are two kinds of open sets in a 1-pt compactification: the old open sets, and sets containing infinity with compact complement. Preimages by $\tilde f$ of the 1st type are open by continuity of the original $f$. Preimages of the 2nd are open by properness of $f$. – Mike F Aug 09 '13 at 17:14
  • @Mike How would you prove that f is proper is equivalent to the limit of |f(t)| going to infinity as t goes to plus or minus infinity? – Keshav Srinivasan Aug 09 '13 at 17:37
  • 1
    Let $D_r$ denoted be the closed disk in $\mathbb{R}^2$ of radius $r >0$, centred on the origin. To show $f:\mathbb{R} \to \mathbb{R}^2$ is proper, one need only check that $f^{-1}(D_r)$ is proper for all $r >0$. This is because any compact $K \subset \mathbb{R}^2$ is contained in in some $D_r$ and then $f^{-1}(K)$ is a closed (since $f$ is continuous) subset of the compact $f^{-1}(D_r)$ and therefore $f^{-1}(K)$ is itself compact. But, what does $f^{-1}(D_r)$ compact mean? That just means it is contained in a bounded interval $[-R,R]$ (since it is already closed by continiuty). – Mike F Aug 09 '13 at 17:56
  • 1
    So, properness is the statement is that, for every $r > 0$, there is an $R > 0$ such that $f(x) \in D_r$ implies $x \in [-R,R]$. Said differently, for every $r > 0$, there is an $R > 0$ such that $|x| > R$ implies $|f(x)| > r$. And that's just the definition of $f$ going to infinity. – Mike F Aug 09 '13 at 17:58
  • @Mike Thanks, that's really clear. If you post it as an answer, I'll be happy to accept it. By the way, do you know how to prove the Jordan curve theorem for a 2-sphere? – Keshav Srinivasan Aug 09 '13 at 18:41
  • 1
    Thanks, but I don't think I have time to type up an answer. Maybe you could collect the details yourself and post them? As for Jordan on the sphere: the idea is to argue a Jordan curve $f: S^1 \to S^2$ is a homeomorphism onto its range (via compactness) and so is not surjective. Deleting some point, or a small disk $D$ if you prefer, outside the range of $f$ from $S^2$ yields a space homeomorphic to $\mathbb{R}^2$ to which the ordinary Jordan theorem applies. Then you argue the deleted region $D$ becomes part of the the "unbounded component" when things are sewn back together. – Mike F Aug 09 '13 at 19:05
  • 1
    @KeshavSrinivasan: If you happen to know a bit algebraic topology, there is a proof in Hatcher's book which is free available in chapter 2.B, proposition 1, a generalization of the Jordan curve theorem. As a special case you get that the complement of an embedding of $S^1$ in $S^2$ has $0$th reduced homology group isomorphic to $\Bbb Z$ which means that it consists of two path-components. It requires some familiarity with homology theory, though. – Stefan Hamcke Aug 09 '13 at 20:55

2 Answers2

11

Consider the curve $\gamma:t\mapsto(e^t,e^{-t}\sin(e^{-t}))$, this maps $\Bbb R$ homeomorphically to the graph $\Gamma$ of the map $x\mapsto \frac1x\cdot\sin\left(\frac1x\right)$. The image of $(-\infty,0)$ is oscillating with increasing amplitude towards $0$, and the image of $(0,\infty)$ is clearly unbounded in $x$-direction.

Define $$C_+=\left\{(x,y)\mid x>0,y>\frac1x\sin\left(\frac1x\right)\right\}\\C_-=\left\{(x,y)\mid x>0,y<\frac1x\sin\left(\frac1x\right)\right\}$$ and $$C_0=(-\infty,0]\times\Bbb R$$

It is easy to prove that all $C'$s, whose union is the complement of $\Gamma$, are path-connected and no point in one of these sets can be joined to a point in another via a path not intersecting $\Gamma$.

On the other hand, since connected components are closed (in $\Bbb R^2-\Gamma$) and $(0,y)\in C_0$ is in the boundary of $C_+$ and $C_-$ (for arbitrary $y$), it follows that $C_-$ and $C_+$ are not the connected components of $\Bbb R^2-\Gamma$. Hence there is only one component in the complement.

Components of $\Bbb R^2-\text{Im}(\gamma)$: $\quad 1$

Path-components: $\qquad\qquad\,\quad 3$

Stefan Hamcke
  • 27,733
1

You can use the single point compactification of $\mathbb{R}^2$, call this $S^2$ to formulate a sufficient criterion for a non-closed curve to divide the plane in two connected components.

If the curve stays unbounded at both ends, i.e. for each $M$ there is an $x_0$ such that if $|x|>x_0$ then $|f(x)| > M $, the curve will cut the plane in two connected halves.

This is equivalent to saying that $f$ can be completed in the one point compactification of ${\mathbb{R}^2}$ which is a sphere, as $f^*$. This is a closed curve.

Take a point not on the image of $f^*$, say $P$ and consider its complement in the sphere $S^2 \setminus P$, this space is homeomorphic to the plane. So, we constructed a simple closed curve in the plane, which has a bounded interior and an unbounded exterior. Because the interior is bounded, this space is disconnected from $P$. So the image of $f^*$ separates $S^2$ into two connected components. These two connected components are exactly the complement of the image of $f$ in $\mathbb{R}^2$, because the point added to $\mathbb{R}^2$ to create $S^2$ is in the image of $f^*$.

My explanation may be a bit chaotic, but there's a valid proof in there.

Answer to the EDIT: Yes that is enough, but the requirement can be weakened quite a bit.

jMdA
  • 1,310