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?

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.
