A related notion of connectedness is this:

Definition. Apathon a topological space X is a continuous map The path is said toconnectx and y in X if f(0)=x and f(1)=y. X is said to bepath-connectedif 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 Pick such that *f*(*x*)=0 and *f*(*y*)=1. There’s a path such that *g*(0)=*x* and *g*(1)=*y*. Now the composition 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 where

To see why it’s not path-connected, suppose is continuous and *f*(0) = (0, 0), *f*(1) = (1, sin(1)).

Let be projection maps to the *x*– and *y*-coordinates respectively. Then contains 0 and 1, so its image is the whole [0, 1] by the intermediate value theorem. Hence, Pick points such that

Since [0, 1] is compact, *f* is uniformly continuous. So there exists ε>0 such that whenever satisfy we have Since [0, 1] is compact, we’ll pick *m*<*n* such that By intermediate value theorem, there’s a point *u* between *t _{m}* and

*t*such that Then but which is a contradiction.

_{n}[ 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 **R**^{2}, 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 is a continuous map of topological spaces and X is path-connected, then so is f(X).

**Proof**.

For any we can pick such that Pick a path such that and Then the composition gives a path which connects to ♦

Proposition 3. If is a collection of path-connected subspaces of X and then so is

**Proof (Sketchy)**.

Pick If then and for some indices *i* and *j*. Since *Y _{i}* and

*Y*are path-connected and contain

_{j}*x*, there’s a path in

*Y*connecting

_{i}*x*to

*y*and a path in

*Y*connecting

_{j}*x*to

*z*. Hence, concatenating gives a path connecting

*y*to

*z*. ♦

Proposition 4. If is a collection of path-connected spaces, then is also path-connected.

**Proof**.

Let Since each is path connected, pick a path such that Now let be the path

To check that *f* is continuous, let’s use the universal property of products. It suffices to show is continuous for each *i*, where is the projection map. But so we’re done. ♦

**Note**

Once again, this fails for the box topology on Since the example 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 *x*, *y* in *X*, a relation *x* ~ *y* 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 thepath-connected components.

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

## Examples

- It’s easy to see that any interval (closed, open or half-open) is path-connected. In particular, it’s connected.
- Hence is a disjoint union of and each of which is a path-connected component.
- The squares [0, 1] × [0, 1] and (0, 1) × (0, 1) are path-connected by proposition 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. - Take the topologist’s sine curve 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.*

Very helpful post.