Path Connectedness

Definition - Path

Let $X$ be a topological space and $p,q\in X$. A path in $X$ from $p$ to $q$ is a continuous map $f:I\to X$ with $f(0)=p$ and $f(1)=q$

A space $X$ is called path-connected if and only if for all $p,q\in X$, there is a path in $X$ from $p$ to $q$

Properties

basically the same as Connectedness.

NOTE

Every continuous image of a path-connected space is path-connected

Proof: Let $p,q\in f(X)$. Take $x\in f^{-1}(p)$ and $y\in f^{-1}(q)$ $f$ is continuous and $\gamma$ is continuous $⟹f\circ \gamma$ is continuous $f\circ \gamma (0)=f(x)=p$ and $f\circ \gamma (1)=f(y)=q$ $⟹f\circ \gamma$ is a path $⟹$ The image $f(X)$ is path-connected

NOTE

Let $\{B_{\alpha }{\}}_{\alpha \in A}$ be a collection of path-connected subspaces of $X$ with a point in common. Then ${\bigcup }_{\alpha \in A}{B}_{\alpha }$ is path-connected

For Proof, see connectedness

relation to Connectedness

NOTE

path-connectedness $⟹$ connectedness

Let $X$ be path-connected. Fix $p\in X$. For any $x\in X$ there exists ${\gamma }_x:p⇝x$ ${\gamma }_x([0,1])$ is connected because $[0,1]$ is connected and ${\gamma }_x$ is continuous.

We can write $X={\bigcup }_{x\in X}{\gamma }_x([0,1])$ Since the union of connected spaces that shares a common point is again connected (point $p$), $X$ must be connected

NOTE

Under manifold, connectedness and path-connectedness are equivalent

Example of the converse direction: Topologist’s sine curve

A standard example is the topologist’s sine curve:

$$S=\{(x,\sin (1/x)):x>0\}\cup \{(0,y):-1\le y\le 1\}\subset {\mathbb{R}}^2.$$

Why this works:

The set $\{(x,\sin (1/x)):x>0\}$ oscillates infinitely often as $x\to 0^{+}$. Its closure adds the whole vertical segment $\{(0,y):-1\le y\le 1\}$. The resulting set $S$ is connected.

But it is not path-connected: there is no continuous path inside $S$ from a point on the vertical segment, such as $(0,0)$, to a point on the oscillating part, such as $(1,\sin 1)$.