## 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 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.

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