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.
thanks for reading.

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 $\infty$.
The above reformulations are easy but the crux of the matter is still in algebraic topology. 

The fact that a Jordan arc does not disconnect the plane is a standard result in algebraic topology, usually proved using homology theory. All texts in algebraic topology have this, but I also like the proof in: Albrecht Dold, A simple proof of the Jordan-Alexander complement theorem, Amer. Math. Monthly 100 (1993), 856-857. A more elementary proof can be found in a Monthly paper by Carsten Thomassen, downloadable from Andrew Ranicki's website. –  Robin Chapman Jul 23 '10 at 9:21

Here is a direct link to the paper The Jordan-Schoenflies Theorem and the Classification of Surfaces  by Carsten Thomassen, which does not use the language of algebraic topology. The result is Proposition 2.11; its proof uses Lemma 2.10, which itself uses Lemma 2.8, which uses Lemmas 2.3 and 2.4... it would be impractical to reproduce it here. 
