# 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?

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What about an open disk? – Etienne Aug 27 '14 at 13:00
The topology on $\mathbb{C}$ is generated by open balls, so any point $z$ is contained in an open ball, which is convex, and hence path-connected. – Travis Aug 27 '14 at 13:00
@Etienne- You have given an example of a path connected component that is open. We are required to prove that every path connected component is open. – Ayush Khaitan Aug 27 '14 at 13:02
@Travis- Yes, but how would you be sure that any such disc lies completely inside the path connected component? – Ayush Khaitan Aug 27 '14 at 13:03
Both of Etienne and Travis are correct, and to see this all you have to do is to apply the definition of the topology on $\mathbb{C}$, to conclude that each $p \in \Omega$ is contained in an open ball $B$ which is itself contained in $\Omega$. – Lee Mosher Aug 27 '14 at 13:04

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.

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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.

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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.

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Actually that's not quite what the question is about. The comments address the actual question. (I stopped myself in the middle of typing the same answer you did). – Lee Mosher Aug 27 '14 at 13:02
@LeeMosher I'm not sure I understand – pppqqq Aug 27 '14 at 13:05