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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?

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What does chain-connected mean? –  Chris Eagle Feb 9 '13 at 23:34
    
x and y are chain connected if: For any epsilon > 0, you can find a finite sequence of points, the first being x and the last y such that every two consecutive points of the sequence are at most at epsilon distance from each other. –  kls3984 Feb 9 '13 at 23:37
    
Your construction could work if you could show the following. Let $M$ be a metric space that is not chainable and $\{A_\lambda\}_{\lambda \in \Lambda}$ the chainable components of $M$. Then there exists $\alpha \in \Lambda$ such that $\inf_{\lambda \in \Lambda \setminus \{\alpha\}} d(A_\lambda,A_\alpha)$ is positive. I say could work, because I'm not sure if its true. –  JSchlather Feb 9 '13 at 23:50
    
Thank you! I think that helped. I'll work on it tomorrow and report back. –  kls3984 Feb 10 '13 at 7:01
    
So I constructed a counterexample to my previous claim as a subspace of a Hilbert space. Perhaps the correct statement is that the there exists a partition $\Lambda=\Gamma \cup \Sigma$ such that $d(A_\Sigma,A_\Gamma)>0$. At any rate my example does not satisfy that. Is this problem from a book? Or is it something you came up with on your own? –  JSchlather Feb 12 '13 at 17:45
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3 Answers

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.

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A shortened version: "Let $X$ be a non-chainable metric space. Then it's non-$\epsilon$-chainable for some $\epsilon>0$. Fix $x$ and let $U$ be the set of points that are $\epsilon$-chain-connected to $x$. Then $(U,X\setminus U)$ is a partition with positive distance." // BTW, I observe from my experience that preceding an answer with something like "I think this works" reduces the number of upvotes that the answer gets. Chances are you'd get upvoted by now if you did not write "I think". –  user53153 Feb 14 '13 at 18:29
    
@5PM Thanks for the advice. I realized after posting that really all you needed to do was like at the $\varepsilon$-chainable components and it falls out, but I didn't have the energy to rewrite. –  JSchlather Feb 14 '13 at 19:25
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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?

share|improve this answer
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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.

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I don't see how this is any different from what the OP was doing. In particular let $X=\{0,1,1/2,1/3,\dots,\}$ and say you get unlucky so that $p=0$. Then your construction fails because while $E=\{0\}$ the other chain connected components get arbitrarily close. –  JSchlather Feb 12 '13 at 15:22
    
Touché, my mistake. I understand the counterexample; I'm working on sorting that out. –  sourisse Feb 12 '13 at 16:54
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