Uniformly path-connected

Definition: A nonempty topological space X is path-connected if for every a,b\in X there exists a continuous map \gamma\colon [0,1]\to X such that \gamma(0)=a and \gamma(1)=b. One usually expresses this by saying that a and b are connected by the (parametrized) curve \gamma.

Yes, I require X to be nonempty. An empty space should not be considered to be path-connected, for the same reason that 1 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 X) 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:

Topologist’s sin

The set S=\{(x,\sin(1/x)\colon 0<x\le 1\} is path-connected, but its closure \overline{S}=S\cup \{(0,y)\colon -1\le y\le 1\} is not.

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

Do not panic. Recall that a continuous function on X does not always extend to a continuous function on \overline{X}, 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 X is uniformly path-connected if there exists a nondecreasing function \omega\colon [0,\infty)\to [0,\infty) such that \omega(0+)=0 and any two points a,b\in X can be connected by a curve of diameter at most \omega(d(a,b)). The diameter of a curve is just the diameter of its image, \mathrm{diam}\,\gamma=\sup\{d(\gamma(t),\gamma(s)\colon t,s\in [0,1]\}. 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 \epsilon\delta 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 X is uniformly path-connected.

Proof. Step 1: for every p\in \overline{X} there exists a curve \gamma\colon [0,1]\colon\overline{X} such that \gamma(0)=p and \gamma(t)\in X for all t>0. Indeed, there is a sequence (x_n) in X that converges to p. For each n=1,2,\dots let \gamma_n\colon [1/(n+1),1/n]\to X be a curve such that \gamma_n(1/(n+1))=x_{n+1}, \gamma_n(1/n)=x_{n}, and \mathrm{diam}\,\gamma_n\le \omega(d(x_{n},x_{n+1}). Define \gamma\colon [0,1]\to \overline{X} by setting \gamma(0)=p and \gamma_{|[1/(n+1),1/n]}=\gamma_n. Let’s check the continuity of \gamma at 0: for all t\in [1/(n+1),1/n] we have d(\gamma(t),p)\le d(x_n,p)+\mathrm{diam}\,\gamma_n\le d(x_n,p)+\omega(d(x_{n},x_{n+1}))\to 0 as n\to\infty. Done.

Step 2. Given distinct points p_1,p_2\in \overline{X}, let \gamma_1, \gamma_2 be curves from Step 1. For j=1,2 pick t_j>0 so that \mathrm{diam}{\gamma_j}_{[0,t_j]}\le d(p_1,p_2). Let q_j=\gamma_j(t_j). Since q_1,q_2\in X, there is a connecting curve \tilde \gamma of diameter at most \omega(d(q_1,q_2))\le \omega(3d(p_1,p_2)). Concatenating \tilde \gamma with the curves {\gamma_j}_{[0,t_j]}, we obtain a curve connecting p_1 to p_2 with diameter at most \omega(3d(p_1,p_2))+2d(p_1,p_2), which we can take as our \tilde \omega for \overline{X}. QED

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

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.