This problem is taken from Stein's text on complex analysis (Exercise 1.5). I was hoping someone can check my proof to see if I have made any mistakes.
Suppose first that $\Omega$ is open and pathwise connected, and that it can be written as $\Omega = \Omega_1 \cup \Omega_2$ where $\Omega_1$ and $\Omega_2$ are disjoint non-empty open sets. Choose two points $w_1 \in \Omega_1$ and $w_2 \in \Omega_2$ and let $\gamma$ denote a curve in $\Omega$ joining $w_1$ to $w_2$. Consider a parameterization $z: [0,1] \rightarrow \Omega$ of this curve with $z(0) = w_1$ and $z(1) = w_2$, and let
$$t^* = \sup_{0 \leq t \leq 1} \{t : z(s) \in \Omega_1, \quad \text{for all } 0 \leq s < t\}$$
Arrive at a contradiction by considering the point $z(t^*)$.
My proof thus far:
First suppose $z(t^*) \in \Omega_1$. If $t^* < 1$, since $\Omega_1$ is open then there exists some $\epsilon > 0$ such that $B_\epsilon(z(t^*)) \subset \Omega_1$. Since $\Omega_1$ is open, this implies $z^{-1}(B_\epsilon(z(t^*)) \subset [0,1]$ and is also open, so it must contain points to the right of $t^*$. This is a contradiction, as these points will map to $\Omega_2$, and so $z(z^{-1}(B_\epsilon(z(t^*))) \not\subset \Omega_1$. If $t^* = 1$, this is also a contradiction as we assumed $z(1) = w_2 \in \Omega_2$. Hence $z(t^*) \not \in \Omega_1$.
Conversely, suppose $t^* \in \Omega_2$. Then since $\Omega_2$ is open, there exists some $\epsilon_2 > 0$ such that $B_{\epsilon_2}(z(t^*)) \subset \Omega_2$. But by continuity, this implies $z^{-1}(B_{\epsilon_2}(z(t^*)))$ contains points to the left of $t^*$, which by definition of $t^*$ must be in $\Omega_1$. Hence this contradicts $z(z^{-1}(B_{\epsilon_2}(z(t^*)))) \subset \Omega_2$, and so $z(t^*) \not \in \Omega^2$.
Therefore, $\Omega$ must be connected.
I am not too confident about taking the pre-image/image of the $\epsilon$-balls to show that there must be points to the right or left of $t^*$.
Best Answer
That's a right idea, but I think, you are complicating it.
Suppose that $z(t^*) \in \Omega_1$ and $t^*=1$. By definition of $t^*$, for every (sufficiently small) $\varepsilon > 0$, $z(t^* - \varepsilon) \in \Omega_1$. However, $z(t^{*})=z(1) = w_2 \in \Omega_2$. Since $\Omega_2$ is open, there is an open ball $B \subseteq \Omega_2$ centered at $z(1)$. By continuity of $z(t)$, for every (sufficiently small) $\varepsilon > 0$, $z(t^* - \varepsilon) \in B \subseteq \Omega_2$, a contradiction with $z(t^* - \varepsilon) \in \Omega_1$.
Suppose that $z(t^*) \in \Omega_1$ and $t^*<1$. There is an open ball $B \subseteq \Omega_1$ centered at $z(t^*)$. Since $B \subseteq \Omega_1$, there must be a $t^{**} > t^*$ for which $z(t^{**}) \in B \subseteq \Omega_1$. This goes against the definition of $t^*$.
Similarly with $z(t^*) \in \Omega_2$, but here always $t^* > 0$ because $\Omega_1$ is open.