Does the Jordan curve theorem apply to non-closed curves? 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?
 A: 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: $\qquad\qquad\,\quad 3$

A: 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.
