Let $\Omega_1$ and $\Omega_2$ are open and connected subsets of the complex plane with $\Omega_1 \cap \Omega_2 \neq \emptyset$. We want to show that $\Omega_1 \cup \Omega_2$ is also connected.
We know that in the complex plane an open set is connected if and only if it is pathwise connected.
Because $\Omega_1$ is pathwise connected we know that for all $z_0,z_1 \in \Omega_1$ there is a path from $z_0$ to $z_1$ in $\Omega_1$. This is the same for the set $\Omega_2$.
Now let $z \in \Omega_1 \cap \Omega_2 $. To proof that $\Omega_1 \cup \Omega_2$ is pathwise connected and thus connected, we want to proof that for every $2$ random picked elements from the set $\Omega_1 \cup \Omega_2$, there exist a path between these elements in $\Omega_1 \cup \Omega_2$. We now look at 2 cases.
Case 1: $z_0,z_1 \in \Omega_1$ or $z_0,z_1 \in \Omega_2$. This case is trivial because we know that these sets are connected.
Case 2: $z_0 \in \Omega_1$ and $z_1 \in \Omega_2$. We know that there exist a path in $\Omega_1$ from $z_0$ to $z$ and also a path in $\Omega_2$ from $z$ to $z_1$. Thus there exists a path in $\Omega_1 \cup \Omega_2$ from $z_0$ to $z_1$ by taking the union of these 2 previous paths. From this we conclude that $\Omega_1 \cup \Omega_2$ is pathwise connected, because $z_0,z_1$ were randomly chosen. And because the set is pathwise connected we know that it is connected.
Can anybody tell me if this is correct of how this could be proved better?
