Math and science::Topology

Path-connectedness. 4 statements

Path-connectedness implies connectedness, the converse is not generally true through.

The topologist's sine curve is an example of a space that is connected but not path-connected. Leinster covers the topologist's sine curve in some detail.

Below, we introduce an iff statement that does hold. It has the form: something ∧ connected ⟺ path-connected. After this, 3 conditions that each imply path-connectedness are presented.

1. Path-connected ⟺ connected and all points have a path-connected neighbourhood

Let $X$ be a topological space. $X$ is path-connected if and only if $X$ is connected and every point of $X$ has at least one path-connected neighbourhood.

2. Corollary. Every connected open subset of $R^{d}$ is path-connected.

3. Path-connectedness and continuous functions

Let $f : X \to Y$ be a continuous map of topological spaces. If $X$ is path-connected, then so is $f X$ .

4. Path-connectedness and products

The product of two path-connected spaces is path-connected.

Proofs

Path-connected ⟺ connected and all points have a path-connected neighbourhood

Let $X$ be a topological space.

Forward implication

Assume that $X$ is path-connected. As shown in a previous card, $X$ is connected as it is path-connected. In addition, we have a path-connected neighbourhood for every point as the set $X$ is a neighbourhood for every point and $X$ is path-connected.

Reverse implication

Assume $X$ is connected and all points have a path-connected neighbourhood. Let $x \in X$ and let $U$ be the set of all elements having a path to $x$ :

U = {y \in X : there exists a path from x to y in X}

Claim: both $U$ and $X ∖ U$ are open in $X$ . Show this in two parts.

$U$ is open: Let $y \in U$ . By assumption, there is an path-connected neighbourhood $W$ of $y$ . Let $w \in W$ . There is a path from $x$ to $w$ as we can join a path from $x$ to $y$ and a path from $y$ to $w$ . Thus, $W \subset U$ . As every element of $U$ has a neighbourhood contained in $U$ , $U$ must be open (A2.9 in Leinster's notes).
$X ∖ U$ is open: Let $y \in X ∖ U$ . By assumption, there is a path-connected neighbourhood $W$ of $y$ . $W \cap U = \emptyset$ , as if there was an element $t \in W \cap U$ we could find a path from $x$ to $y$ by joining the path from $x$ to $t$ and the path from $t$ to $y$ ; but such a path would contradict $y \notin U$ . We can now state that $W \in X ∖ U$ , as $W \subseteq X$ and $W \cap U = \emptyset$ . As every element of $X ∖ U$ has a neighbourhood contained in $X ∖ U$ , $X ∖ U$ must be open (again by A2.9).

So, we have both $U$ and $X ∖ U$ being open. However, $X$ is connected, so one of the sets must be empty and the other must equal $X$ . $x$ itself is in $U$ , so $U = X$ . $U$ is path-connected by definition, so $X$ is path-connected.

Context

Source

Leinster, p79