Let $\Omega\subset \Bbb{C}$ be an open set. My textbook states that every path connected component of $\Omega$ is open.
I can't seem to understand why that is. Why does every point have to contained in a path-connected neighbourhood which lies entirely inside the path connected component?
Best Answer
Let $P$ be any path component of $\Omega$. Let $p \in P$ be any point. Since $\Omega$ is open, there exists an open ball $B$ such that $p \in B$ and $B \subset \Omega$. Since $B$ is path connected, it must be contained in $P$, by definition of path component. Therefore $P$ is open.