Equivalent characterization of chain connectedness of a metric space 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?
 A: 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. 
A: It seems like you have the right idea with proving the contrapositive. That is, suppose $X$ is not chain connected. We want to show there's a uniformly equicontinuous family which is pointwise bounded at some point of $X$ but is not pointwise bounded at some other point of $X$. 
To this end, let $p\in X$ and define $E=\{x \in X : x \sim p\}$, where $x \sim y \iff \forall \delta >0$ there is a $\delta$-chain from $x$ to $y$. If $E=X$, then $X$ is chain connected, so by hypothesis $X\setminus E$ is nonempty. Let $$f_n(x) = \left\{
     \begin{array}{ll}
      n & \,\, x \notin E\\
      0 & \,\, x \in E
     \end{array}
   \right.
$$
Notice that $f_n(p)=0$ for all $n\ge 1$, so pointwise boundedness of the family at $p$ is clear.
We show uniform equicontinuity of $(f_n)_{n\ge 1}$. Let $\epsilon >0$. We need a $\delta>0$ such that for any $f_n$ and any $x,y \in X$, the expression $|f_n(x)-f_n(y)|$ is less than $\epsilon$ whenever $d(x,y)<\delta$. But the former expression is one of three things:


*

*$|n-n|=0$ if $x,y \notin E$

*$|0-0|=0$ if $x,y \in E$

*$|n-0|=n$ if only one of $x$ and $y$ is in $E$.


The last case never happens though, because if $d(x,y)<\delta$, then $x$ and $y$ are in the same equivalence class. So $\delta$ can be any positive number, and $(f_n)_{n\ge 1}$ is uniformly equicontinuous and pointwise bounded at $p$.
However, $\{f_n(q)\}_{n\ge 1}$ is not bounded at $q\in X\setminus E$, because $f_n(q) = n$ for all $n \in \mathbb{N}$. So we've proved the contrapositive.
A: Apologies. I should have logged in.
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?
