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.


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. ♦

warningThe 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).


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.


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. ♦


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.


  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.
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1 Response to Topology: Path-Connected Spaces

  1. Nkechi Nnadi says:

    Very helpful post.

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