TITLE: Equivalent characterization of chain connectedness of a metric space
QUESTION [6 upvotes]: I'm having difficulty with proof. It is that the following is an equivalent characterization of chain connectedness for a metric space $M$:
Point-wise boundedness at a point of an equicontinuous family of functions (from M to the real numbers) implies pointwise boundedness at all points.
Pointwise boundedness means boundedness of the set of values of all functions at a given point.
Now, my attempt was to construct an equicontinuous family of functions that could be useful. For instance, you can take $A$ be the set of all points chain connected to the point $a\in M$. Then you can take $f_n$ defined as $0$ on $A$ and $n$ otherwise. The functions $f_n$ are bounded at $a$, so equicontinuity would give us the desired result. The challenge I'm facing is showing equicontinuity. Plus, it does not seem like a good choice of functions. For instance, take $$M=\{r\in\mathbb R\, |\, r<0, r=0 \text{ or } r=1/n \text{ for all }n \text{  natural}\}.$$ Use the standard Euclidean metric on $\mathbb R$. Take $a=0$. It seems like you can find $x$ and $y$ arbitrarily close such that $f_n(x)$ and $f_n(y)$ still differ by $n$.
Thank you!

So I think the example I gave shows that the distances between A and the rest of the chain connected components can have a zero infimum, which is what makes me skeptical that this will work. So do you think the choice functions is bad?
EDIT: Could a better choice of A, i.e. a better choice of a, be helpful?

REPLY [2 votes]: Okay, I think this is it. Let $X$ be a metric space. Then $X$ is chainable if and only if for any partition $U \cup Y= X$ such that $U,Y \neq \emptyset$ we have $d(U,Y)=0$. The first implication is immediate, so we need only show that a space $X$ which is not chainable has such a partition. Let $X$ be a non-chainable metric space. Then we have two points $x,y \in X$  $\newcommand{\eps}{\varepsilon}$ such that there is not an $\eps$-chain connecting $x$ and $y$. Let $A_y, A_x$ be the $\eps$-chain components of $y$ and $x$. By assumption $A_y \cap A_x=\emptyset$ and $d(A_y,A_x)\geq \eps$. Let $V=X \setminus (A_y,A_x)$ if $V$ is empty we are done. Otherwise note that $d(V,A_x),d(V,A_y)\geq \eps$ in particular $d(V,A_x\cup A_y)\geq \eps$ thereby $V, A_x \cup A_y$ is such a partition.
Your counterexample will now work by taking such a partition and defining on the two subsets. The key idea here is $\eps$-chainable instead of chainable, it gives much more control.