# Continuous curve, traps itself outside the unit circle.

Lets say i have an injective continuous curve $\sigma$ in $\mathbb{C}$, indexed on $[0,\infty)$ and converging to $\infty$. If $\vert \sigma(0)\vert>0$ , is it possible that it can trap itself outside the unit circle? By that i mean, that there doesn't exist an extension of the curve, so that the beginning point is on the unit circle? My intuitive guess if of course no, but i wonder if there is a simple proof that doesn't require more than the first course in topology . I would also appreciate if someone could confirm that my guess is correct.

Let's compactify $\mathbb C$ to $S^2=\mathbb C\cup\{\infty\}$, and add the point at $\infty$ to the image of $\sigma$. Your question amounts to whether the complement of a simple arc $\gamma$ in $S^2$ is path-connected. (A simple arc is a homeomorphic image of $[0,1]$). This is equivalent to asking whether the complement of a simple arc in $\mathbb R^2$ is connected, because we can apply a Möbius transformation to $S^2$ to move some point of the complement of $\gamma$ to $\infty$.