Is path-connectedness same as connectedness in the complex plane?

Question

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?

Answer 1

These are the same thing when $D$ is an open subset of $\mathbb{C}$, or more generally, of any topological space $X$ which is locally path connected. This means that for every point $x \in X$ and every open set $U \ni x$, there is an open set $V$ with $x \in V \subseteq U$ and $V$ path-connected. It's easy to see that this property holds for $\mathbb{C}$ or any other manifold, since every open set contains arbitrarily small open balls centered at any point.

In complete generality, any path-connected space is connected, so we need to show the converse. (This should be proved in any reference on general topology, for example the book Topology by Munkres).

Let $X$ be a locally path-connected space and $D \subseteq X$ connected and open (and non-empty). Let $x\in D$ be an arbitrary point. We may define $D_x$ to be the set of points $z \in D$ such that there is a path connecting $x$ to $z$. It is open: if $z \in D_x$, then take some path-connected open set $U$ around $z$ which is contained in $D$. Since we have a path in $D$ from $x$ to $z$ and from $z$ to $z'$ for any $z' \in U$, we have a path in $D$ from $x$ to $z'$ for all $z' \in U$, and thus $U \subseteq D_x$, so $D_x$ is open. Now, it suffices to show that $D \setminus D_x$ is open, since if $D_x$ were both open and closed, it would have to be equal to $D$ since $D$ is connected (note that $x \in D_x$, so it's non-empty). To do this, let $z \in D$ be such that no path in $D$ connects $z$ to $x$. Choose some path-connected open set $U$ with $z \in U \subseteq D$. We want to show that $U \subseteq D \setminus D_x$. Thus, assume there is some $y \in D_x \cap U$. Then, there is a path connecting $x$ to $y$. But since $U$ is path-connected, there is a path connecting $y$ to $z$, and thus by combining these paths we get a path from $x$ to $z$, a contradiction.

Note It's fairly common to ignore the difference between the two concepts in contexts other than general topology, since many of the topological spaces which are important to people in other fields are always locally path-connected (e.g. manifolds). However, these concepts can differ, as shown in the example of the topologist's sine curve.

Answer 2

For open subsets $S \subseteq \mathbf C$ connectedness and path-conectedness coincide, due to their local-pathconnectedness:

Lemma. Let $X$ be a locally pathwise connected and connected topological space. Then $X$ is pathwise connected.

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.