## Topology: Path-Connected Spaces

A related notion of connectedness is this:

Definition. A path on a topological space X is a continuous map $f:[0, 1] \to X.$ The path is said to connect x and y in X if f(0)=x and f(1)=y. X is said to be path-connected if any two points can be connected by a path.

In a sense, path-connectedness is more active since one requires an explicit path to establish it, while the earlier connectedness is more passive since it simply indicates a failure to decompose as a topological disjoint union. The two are obviously related, starting from:

Theorem 1. A path-connected space X is connected.

Proof.

Suppose X is path-connected but not connected. There’s a surjective continuous map $f:X\to \{0,1\}.$ Pick $x, y\in X$ such that f(x)=0 and f(y)=1. There’s a path $g:[0, 1] \to X$ such that g(0)=x and g(1)=y. Now the composition $f\circ g:[0, 1] \to\{0,1\}$ is continuous and surjective, which contradicts the fact that [0, 1] is connected. ♦

The converse is not true: the topologist’s sine curve is connected but not path-connected. Let $X = Y\cup Z \subset \mathbf{R}^2,$ where $Y = \{0\}\times [0, 1],$ $Z=\{(x, \sin(1/x)) : 0 < x \le 1\}.$

To see why it’s not path-connected, suppose $f:[0, 1] \to X$ is continuous and f(0) = (0, 0), f(1) = (1, sin(1)).

Let $\pi_x, \pi_y : X\to \mathbf{R}$ be projection maps to the x– and y-coordinates respectively. Then $\pi_x \circ f:[0, 1]\to \mathbf{R}$ contains 0 and 1, so its image is the whole [0, 1] by the intermediate value theorem. Hence, $\text{im}(f)\supset Z.$ Pick points $t_0, t_1, t_2, \ldots \in [0, 1]$ such that $f(t_n) = ((2n\pi + \frac \pi 2)^{-1}, 1).$

Since [0, 1] is compact, f is uniformly continuous. So there exists ε>0 such that whenever $t, u\in [0, 1]$ satisfy $|t-u|<\epsilon,$ we have $|\pi_y(f(t)) - \pi_y(f(u))| <1.$ Since [0, 1] is compact, we’ll pick m<n such that $|t_m - t_n|<\epsilon.$ By intermediate value theorem, there’s a point u between tm and tn such that $\pi_x(f(u)) = ((2m\pi + \frac {3\pi}2)^{-1}, -1).$ Then $|t_m - u| < \epsilon$ but $|\pi_y(f(t_m)) - \pi_y(f(u))| = |1-(-1)|=2>1$ which is a contradiction.

[ Notice it took quite a bit of effort to prove a seemingly obvious claim, and we needed compactness to prove it. ]

Note also that X is closed in R2, and every point in Y is a point of accumulation of Z, so cl(Z) = X. In short, we have the first bummer.

Conclusion. The closure of a path-connected subset Y of X is not necessarily path-connected.

Thankfully, the next result does carry over.

Proposition 2. If $f:X\to Y$ is a continuous map of topological spaces and X is path-connected, then so is f(X).

Proof.

For any $y_0, y_1\in f(X)$ we can pick $x_0, x_1 \in X$ such that $f(x_0) = y_0, f(x_1) = y_1.$ Pick a path $f:[0, 1] \to X$ such that $f(0)=x_0$ and $f(1)=x_1.$ Then the composition gives a path $g\circ f:[0, 1] \to Y$ which connects $y_0$ to $y_1.$ ♦

Proposition 3. If $\{Y_i\}$ is a collection of path-connected subspaces of X and $\cap_i Y_i\ne\emptyset,$ then so is $Y:= \cup_i Y_i.$

Proof (Sketchy).

Pick $x\in \cap_i Y_i.$ If $y,z\in Y,$ then $y\in Y_i$ and $z\in Y_j$ for some indices i and j. Since Yi and Yj are path-connected and contain x, there’s a path in Yi connecting x to y and a path in Yj connecting x to z. Hence, concatenating gives a path connecting y to z. ♦

Proposition 4. If $\{X_i\}$ is a collection of path-connected spaces, then $X:=\prod_i X_i$ is also path-connected.

Proof.

Let $(x_i), (y_i)\in X.$ Since each $X_i$ is path connected, pick a path $f_i :[0,1]\to X_i$ such that $f_i(0) = x_i, f_i(1) = y_i.$ Now let $f:[0, 1]\to X$ be the path $f(t) := (f_i(t))_i \in X.$

To check that f is continuous, let’s use the universal property of products. It suffices to show $\pi_i \circ f : [0,1]\to X_i$ is continuous for each i, where $\pi_i :X\to X_i$ is the projection map. But $\pi_i \circ f = f_i,$ so we’re done. ♦

Note

Once again, this fails for the box topology on $\prod_i X_i.$ Since the example $\mathbf{R}^\mathbf{N}$ in the previous article is not connected, it cannot be path-connected.

## Path-Connected Components

As before, we obtain the concept of path-connected components. We define, for points xy in X, a relation xy if and only if they belong to some path-connected component. Proposition 3 then tells us this gives an equivalence relation.

Definition. The equivalence classes of the above-mentioned relation are called the path-connected components.

Since a path-connected component is automatically connected, each connected component is a disjoint union of path-connected components.

## Examples

1. It’s easy to see that any interval (closed, open or half-open) is path-connected. In particular, it’s connected.
2. Hence $X = [0, 1) \cup (2, 3]$ is a disjoint union of $[0, 1)$ and $(2, 3],$ each of which is a path-connected component.
3. The squares [0, 1] × [0, 1] and (0, 1) × (0, 1) are path-connected by proposition 4.
4. Consider Q as a subspace of R. Since the connected components are singleton sets, the path-connected components can’t break them down any further.
5. Take the topologist’s sine curve $X=Y\cup Z$ above. Y and Z are both path-connected since they’re homeomorphic to intervals. Since X is not path-connected, the path-connected components must be Y and Z. Note that Y is open while Z is closed in X. This is one example where connected components decompose further into path-connected components.
This entry was posted in Notes and tagged , , , , , . Bookmark the permalink.