I stumbled upon the following problem in one of my problem sets on the Riemann Mapping Theorem, which states: Let $U \subseteq \mathbb{C}$ be simply connected and open. Then, for every point $\alpha$ in the boundary of $U$, there exists a curve $\gamma :[0,1] \to \mathbb{C} \cup \{ \infty \} $ (by curve we will understand continuous function) that joins $\alpha$ and $\infty$, that is to say, $\gamma(0) = \alpha$ and $\gamma(1) = \infty$ and such that $U \subset \mathbb{C} \setminus \gamma ([0,1])$. This proposition was used in proving the following lemma used in the proof of the Riemann Mapping theorem:
If $U \subset \mathbb{C}$, $U \neq \mathbb{C}$ is simply connected and open, then there exists a bijective analytic mapping $f: U \to V$ such that $V$ is an open subset of $\{ z \in \mathbb{C} : |z| = 1\}$.
However, I have tried to prove the theorem by contradiction and obtaining a closed curve in $U$ whose interior is not totally contained in $U$, and some other attempts, but, I have not obtained much from them. I need some help in seeing how to prove this result.
The characterization: $G \subseteq \mathbb{C}$ is simply connected iff for every closed jordan curve $\gamma : [0,1] \to G$ its interior lies entirely in $G$ can be used.
The course which I am attending to and from where I got this problem is mainly based in Lang's and Silvermann's texbooks on the subject, but, I did not found any information concerning the result in them.
Addendum: I have been able to prove that if $G$ is open and simply connected then its complement in the Riemann Sphere or in the extended plane, must be connected, however, I have not been able to prove that its complement is simply connected
Best Answer
Let $Z\subset {\mathbb C}$ denote the topologist's sine curve, i.e. the union $X\cup Y$ of $$ X=\{(x, \sin(1/x)): x>0\} $$ and the segment $Y=\{(0,y): -1\le y\le 1\}$. Then $Z$ is connected but not path-connected: One cannot connect by a path in $Z$ any point of $X$ to any point of $Y$. Set $\hat{Z}:= Z\cup \{\infty\}$, where $\infty$ is the point at infinity of the Riemann sphere ${\mathbb C}\cup \{\infty\}$. The same argument shows that there is no path in $\hat{Z}$ connecting a point in $Y$ to $\infty$ (it is a nice exercise in analysis to work out details). Since $Z$ is connected, its complement, $U= {\mathbb C}\setminus Z$ is simply-connected.
Thus, $U$ provides a counter-example to the claim from the notes you are trying to prove.