In the book Complex Variables and Applications by Churchill page 32:
An open set S is connected if each pair of points $z_1$ and $z_2$ in it can be joined by a polygonal line, consisting of a finite number of line segments joined end to end, that lies entirely in S.
This is the dentition of path-connectedness in General Topology not connectedness. Does the mentioned text implies that in the complex plane path-connectedness is same as connectedness? or it's because of different terminology in different subjects?
Best Answer
For open subsets $S \subseteq \mathbf C$ connectedness and path-conectedness coincide, due to their local-pathconnectedness:
Proof. Let $x\in X$. Define $$ A := \{y \in X : \text{There is a path in $X$ joining $x$ and $y$}\} $$ As $x \in A$, $A \ne \emptyset$. If $y \in A$, then as $X$ is locally pathwise connected, there is a pathwise connected neighbourhood $U$ of $y$. Then, by joining paths, we see that $U \subseteq A$. Hence $A$ is open. If $y\in X \setminus A$, no point of a pathwise connected neighbourhood of $y$ can belong to $A$. Hence $A$ is closed. As $A$ is clopen and non-empty, the connectedness of $X$ implies $A = X$. So $X$ is pathwise connected.