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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?

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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. – Matthew Pressland Aug 9 '13 at 15:20
@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 9 '13 at 15:56
@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 9 '13 at 16:03
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 9 '13 at 16:22
@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 9 '13 at 17:01

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: $\qquad\qquad\,\quad 3$

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Just because they're path components doesn't mean they're components. You need a bit more. – dfeuer Aug 9 '13 at 16:11
@dfeuer: Thanks, I think there's even only one component here. – Stefan Hamcke Aug 9 '13 at 16:15
@StefanH.: Interesting. There's more here than meets the eye. – Nate Eldredge Aug 9 '13 at 17:05

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