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

Question

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?

Answer 1

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.

Answer 2

Let $\Omega_p$ be the path component of $\Omega$ which contains the point $p$ and let $x$ be any point in $\Omega_p$. Let $\epsilon$ be such that $B_{\epsilon}(x)\subset\Omega$ which must exists because $\Omega$ is open. Note that $B_{\epsilon}(x)$ is an open disk and so is path connected. In particular for any $y\in B_{\epsilon}(x)$, the element $y$ must be in the same path component of $\Omega$ as $x$, and therefore $p$, because path-connectedness is an equivalence relation. It follows that $B_{\epsilon}(x)\subset\Omega_p$ and so $\Omega_p$ is open.

Answer 3

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.