**Definition**: A nonempty topological space is *path-connected* if for every there exists a continuous map such that and . One usually expresses this by saying that and are connected by the (parametrized) curve .

Yes, I require to be nonempty. An empty space should not be considered to be path-connected, for the same reason that is not considered to be a prime number. Similar to the uniqueness of prime factorization, every topological space can be written as a disjoint union of path-connected spaces (called the path-components of ) in a unique way.

The definition of path-connectedness is much more intuitive that the notion of connectedness as the absence of nonempty proper closed-open subsets. Yet it has a problem: the closure of a path-connected subset is not necessarily path-connected. An obligatory picture:

The set is path-connected, but its closure is not.

This is where path-connectedness loses compared to connectedness. Indeed, it is hard to do much on a metric space that is not complete, so we usually pass to its completion . But we would not want to lose any geometric features of in this process. And the path-connectedness may be lost, as in the above example.

Do not panic. Recall that a continuous function on does not always extend to a continuous function on , but a uniformly continuous function does. What we need is a notion of *uniformly path-connected.*

Following the ideology of my earlier post, I will say that is **uniformly path-connected** if there exists a nondecreasing function such that and any two points can be connected by a curve of diameter at most . The diameter of a curve is just the diameter of its image, . One can imagine using length instead of diameter, but this would be a more restrictive definition ruling out all the beautiful snowflakes.

Had I defined “uniformly path-connected” along the lines of the traditional – definition of uniform continuity, “uniformly path-connected” would not imply “path-connected”. Another reason why I’m saying that the definition of uniform continuity is wrong.

**Claim.** the closure of a uniformly path-connected set is uniformly path-connected.

**Proof.** Step 1: for every there exists a curve such that and for all . Indeed, there is a sequence in that converges to . For each let be a curve such that , , and . Define by setting and . Let’s check the continuity of at : for all we have as . Done.

Step 2. Given distinct points , let , be curves from Step 1. For pick so that . Let . Since , there is a connecting curve of diameter at most . Concatenating with the curves , we obtain a curve connecting to with diameter at most , which we can take as our for . **QED**

I could play with string/nonstrict inequalities to make , but the post is boring enough already.