I need to show that if (X,p) is a non-compact metric space, then there exists a metric p* equivalent to p such that (X,p*) is not complete.

I greatly appreciate your help!

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- Jul 10th 2009, 01:56 PMnubmathieChallenging Problem!!! (Equivalent Metrics)
I need to show that if (X,p) is a non-compact metric space, then there exists a metric p* equivalent to p such that (X,p*) is not complete.

I greatly appreciate your help! - Jul 10th 2009, 08:38 PMJose27
What is your definition of equivalent metrics? Mine is that they generate the same topology, but with that take an infinite set $\displaystyle X$ with the discrete metric then for any equivalent metric we have that for every $\displaystyle x \in X$ there exists $\displaystyle a_x \in \mathbb{R} ^+$ such that $\displaystyle B_X(x,a_x):= \{ y \in X : d(x,y)<a_x \} =\{ x \}$ and so every Cauchy sequence in any equivalent metric is eventually constant and so it converges, therefore X is complete with any metric equivalent to the discrete metric.

- Jul 10th 2009, 09:15 PMnubmathie
You are right about the definition of equivalent metrics. But I'm trying to prove the statement for any metric p, not just the discrete metric.

- Jul 10th 2009, 09:42 PMJose27
- Jul 11th 2009, 03:40 AMLaurent
I can't agree with your counter-example; I think you should try to explicitate the part in red to see why it fails.

Consider the metric on $\displaystyle \mathbb{N}$ defined by $\displaystyle d(n,n)=0$, $\displaystyle d(n,n+1)=\frac{1}{2^n}$ and, for $\displaystyle m\leq n$, $\displaystyle d(m,n)=\sum_{k=m}^{n-1} d(k,k+1)$. (This example corresponds to considering the subset $\displaystyle \{\frac{1}{2^n}|n\geq -1\}$ with the usual topology of $\displaystyle \mathbb{R}$). Since $\displaystyle B(n,\frac{1}{2^n})=\{n\}$ (open ball), this defines the discrete topology. Yet you can see that the sequence $\displaystyle (n)_{n\geq 0}$ is a Cauchy sequence that doesn't converge. - Jul 11th 2009, 06:58 AMhalbard
Isn't completeness a property of metrics rather than of topologies?

For example, take $\displaystyle \mathbb R$ with the usual metric $\displaystyle d(x,y)=|x-y|$. It is readily shown that $\displaystyle d$ is equivalent to the metric $\displaystyle \bar d$ given by

$\displaystyle \bar d(x,y)=\left|\frac x{1+|x|}-\frac y{1+|y|}\right|$.

Now $\displaystyle \mathbb R$ is $\displaystyle d$-complete but not $\displaystyle \bar d$-complete since $\displaystyle (n)_{n\in\mathbb N}$ is a $\displaystyle \bar d$-Cauchy sequence that does not converge to anything in $\displaystyle (\mathbb R,\bar d)$. - Jul 11th 2009, 07:47 AMLaurent
- Jul 11th 2009, 09:09 AMnubmathie
Laurent's example is exactly the one I've thought of. Now we need to extend it to any compact metric spaces.

The idea is to assume WLOG that (X,p) is complete but not totally bounded. So we can take an unbounded sequence in (X,p), and make it Cauchy under p* by somehow contracting the distances. Since it was unbounded under p, intuitively it will be non-convergent under p*.

Or any other ideas? - Jul 14th 2009, 11:47 AMJose27
...Ouch.

Well, I was thinking of using something like Alexandroff's compactification by a point. Like when you identify $\displaystyle \mathbb{R} ^n$ with $\displaystyle A:= \mathbb{S} ^n - \{ e_n \}$ where $\displaystyle \mathbb{S} ^n := \{x \in \mathbb{R} ^{n+1} : \Vert x \Vert =1 \}$ and any sequence converging to $\displaystyle \infty$ (or $\displaystyle e_n$) is a Cauchy sequence in $\displaystyle A$ that doesn't converge in $\displaystyle \mathbb{R} ^n$. So far no luck in generalizing though. Is there a natural metric on your compactification if it is made over a metric space? - Jul 14th 2009, 11:47 PMOpalg
I don't know of any straightforward explicit construction of a metric on the one-point compactification of a metric space, but you could use a metrization theorem to show that one must exist. The restriction of this metric to the original space would then provide a solution to the problem.

- Jul 15th 2009, 03:14 AMLaurent
This construction based on the one-point compactification will indeed work, but under some conditions. For instance the one-point compactification is not always Hausdorff (the base space needs to be locally compact for that) hence not metrizable. And Urysohn's theorem applies to second-countable spaces.

By browsing through the Wikipedia, it seems that Stone-Cech compactification (a many-points compactification) gives a Hausdorff compact space, but Urysohn's theorem probably doesn't apply... I haven't other clues anyway. - Jul 15th 2009, 05:00 AMOpalg
- Jul 15th 2009, 09:34 AMJose27
Okay, but the topology we induce on $\displaystyle X \bigcup \{ p \}$ is not necessarily that of the ususal compactification: For example when we do the stereographic projection of $\displaystyle \mathbb{S} ^n$ on $\displaystyle \mathbb{R} ^n$ the topology we give to $\displaystyle \mathbb{S} ^n$ is $\displaystyle \tau= \{ U \subset X:U$ is open in $\displaystyle X \}$ $\displaystyle \bigcup \{ p \in U : U=(X \bigcup \{ p \} )-V$ where $\displaystyle V \subset X$ is compact $\displaystyle \}$ and $\displaystyle X= \mathbb{R} ^n$.

- Jan 23rd 2010, 10:23 AMns1954
Two metrics $\displaystyle d_1$ and $\displaystyle d_2$ on same X are equivalent, if they produces same topologies, i.e. $\displaystyle {\tau}_{d_1}={\tau}_{d_2}$.

For general case, if the staments is true, remind that if $\displaystyle (X,d)$ isn't compact, he contain sequence $\displaystyle {\{x_n\}}_{n=1}^{\infty}$ with no convergent subsequences, but he must not be Cashy. So, question is here how to use this to construct equivalent non-complete metric in term of this sequences.