Openness of path connected components of open subsets of $\mathbb C$ 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?

It is because $\mathbb C = \mathbb R ^2$ is locally path connected, and so every open subset of $\mathbb C$. For a locally path connected topological space, the path-connected components are open and coincide with the connected components.
In fact, let $X$ be a topological space which is locally path connected and $C\subset X$ a path connected. If $p \in C$, there's a neighbourood $U$ of $p$ which is path connected. Since being connected by a path is an equivalence relation, every point of $U$ is also a point of $C$ which says precisely that $C$ is open.
