Path Connectedness

Path connectedness is a stronger form of connectedness defined in terms of the existence of continuous paths between points. While every path-connected space is connected, the converse fails in general. Path connectedness is the more intuitive notion and is the foundation of homotopy theory.

Definition

Definition4.4Path

Let $X$ be a topological space. A path in $X$ from $x$ to $y$ is a continuous map $\gamma: [0, 1] \to X$ with $\gamma(0) = x$ and $\gamma(1) = y$ . We say $x$ and $y$ are the endpoints of the path.

Definition4.5Path-Connected Space

A topological space $X$ is path-connected if for every pair of points $x, y \in X$ , there exists a path in $X$ from $x$ to $y$ .

ExamplePath-Connected Spaces

$\mathbb{R}^n$ is path-connected: The straight-line path $\gamma(t) = (1-t)x + ty$ connects any two points.
$S^n$ for $n \geq 1$ is path-connected: Any two non-antipodal points are connected by a great circle arc. Antipodal points are connected by going through a third point.
Convex sets: Any convex subset of $\mathbb{R}^n$ is path-connected.
The punctured plane $\mathbb{R}^2 \setminus \{0\}$ is path-connected (but $\mathbb{R} \setminus \{0\}$ is not).

Path Connectedness Implies Connectedness

Theorem4.3Path-Connected Implies Connected

Every path-connected space is connected.

Proof

Suppose $X$ is path-connected and $X = U \cup V$ is a separation. Pick $x \in U$ and $y \in V$ . Since $X$ is path-connected, there exists a continuous path $\gamma: [0,1] \to X$ with $\gamma(0) = x$ and $\gamma(1) = y$ .

Then $[0,1] = \gamma^{-1}(U) \cup \gamma^{-1}(V)$ is a separation of $[0,1]$ : both sets are open (since $\gamma$ is continuous), nonempty ( $0 \in \gamma^{-1}(U)$ , $1 \in \gamma^{-1}(V)$ ), and disjoint. But $[0,1]$ is connected, a contradiction.

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RemarkThe Converse Fails

The topologist's sine curve $S = \overline{\{(x, \sin(1/x)) : x > 0\}}$ is connected but not path-connected. There is no continuous path from $(0, 0)$ to $(1/\pi, 0)$ that stays in $S$ . Intuitively, any path approaching the origin must oscillate infinitely, preventing continuity.

Path Components

Definition4.6Path Component

The relation "there exists a path from $x$ to $y$ " is an equivalence relation on $X$ (using constant paths for reflexivity, reversed paths for symmetry, and concatenated paths for transitivity). The equivalence classes are called path components of $X$ .

The set of path components is denoted $\pi_0(X)$ .

ExamplePath Components

$\pi_0(\mathbb{R}^n) = \{*\}$ (one path component) for $n \geq 1$ .
$\pi_0(\mathbb{R} \setminus \{0\}) = \{(-\infty, 0), (0, \infty)\}$ (two path components).
$\pi_0(\mathbb{Q}) = \mathbb{Q}$ (every point is its own path component, since $\mathbb{Q}$ is totally disconnected).
The topologist's sine curve has one connected component but two path components.

Path Connectedness under Standard Operations

Theorem4.4Preservation of Path Connectedness

The continuous image of a path-connected space is path-connected.
If $\{A_\alpha\}$ is a family of path-connected subsets with $\bigcap_\alpha A_\alpha \neq \emptyset$ , then $\bigcup_\alpha A_\alpha$ is path-connected.
A product $\prod_\alpha X_\alpha$ is path-connected if and only if each $X_\alpha$ is path-connected.

Proof

(1): Let $f: X \to Y$ be continuous with $X$ path-connected. For $y_1, y_2 \in f(X)$ , choose $x_1, x_2$ with $f(x_i) = y_i$ . Let $\gamma$ be a path from $x_1$ to $x_2$ in $X$ . Then $f \circ \gamma$ is a path from $y_1$ to $y_2$ in $f(X)$ .

(2): Fix $p \in \bigcap_\alpha A_\alpha$ . For any $x, y \in \bigcup_\alpha A_\alpha$ , $x \in A_\alpha$ and $y \in A_\beta$ for some $\alpha, \beta$ . There are paths from $x$ to $p$ in $A_\alpha$ and from $p$ to $y$ in $A_\beta$ . Concatenating gives a path from $x$ to $y$ .

(3): A path in $\prod X_\alpha$ is a map $\gamma: [0,1] \to \prod X_\alpha$ , which is continuous if and only if each $\pi_\alpha \circ \gamma: [0,1] \to X_\alpha$ is continuous (by the universal property of products). So $\gamma$ is a path from $\mathbf{x}$ to $\mathbf{y}$ in the product if and only if each $\pi_\alpha \circ \gamma$ is a path from $x_\alpha$ to $y_\alpha$ in $X_\alpha$ .

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Locally Path-Connected Spaces

Definition4.7Locally Path-Connected

A space $X$ is locally path-connected if for every $x \in X$ and every open neighborhood $U$ of $x$ , there exists an open neighborhood $V$ of $x$ with $V \subseteq U$ such that $V$ is path-connected.

Equivalently, $X$ has a basis of path-connected open sets.

Theorem4.5Locally Path-Connected and Connected Implies Path-Connected

If $X$ is connected and locally path-connected, then $X$ is path-connected.

Proof

Fix $x_0 \in X$ . Let $U$ be the set of points that can be connected to $x_0$ by a path in $X$ . We show $U$ is clopen.

$U$ is open: If $x \in U$ , let $V$ be a path-connected neighborhood of $x$ . For any $y \in V$ , concatenate a path from $x_0$ to $x$ with a path from $x$ to $y$ in $V$ . So $V \subseteq U$ .

$U$ is closed: If $x \in \overline{U}$ , let $V$ be a path-connected neighborhood of $x$ . Then $V \cap U \neq \emptyset$ ; pick $y \in V \cap U$ . There is a path from $x_0$ to $y$ , and a path from $y$ to $x$ in $V$ . Concatenating gives a path from $x_0$ to $x$ , so $x \in U$ .

Since $U$ is clopen, nonempty ( $x_0 \in U$ ), and $X$ is connected, $U = X$ .

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ExampleLocally Path-Connected Spaces

Open subsets of $\mathbb{R}^n$ are locally path-connected.
Manifolds are locally path-connected (locally homeomorphic to $\mathbb{R}^n$ ).
CW complexes are locally path-connected.
The topologist's sine curve is not locally path-connected at points on the segment $\{0\} \times [-1, 1]$ .