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Math Help - Challenging Problem!!! (Equivalent Metrics)

  1. #1
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    Challenging 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!
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    What is your definition of equivalent metrics? Mine is that they generate the same topology, but with that take an infinite set X with the discrete metric then for any equivalent metric we have that for every x \in X there exists a_x \in \mathbb{R} ^+ such that 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.
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    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.
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    Quote Originally Posted by nubmathie View Post
    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.
    Actually, I think what I did was show a counterexample: Any infinite discrete space is not compact, but with any equivalent metric is complete.
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  5. #5
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    Quote Originally Posted by Jose27 View Post
    What is your definition of equivalent metrics? Mine is that they generate the same topology, but with that take an infinite set X with the discrete metric then for any equivalent metric we have that for every x \in X there exists a_x \in \mathbb{R} ^+ such that 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.
    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 \mathbb{N} defined by d(n,n)=0, d(n,n+1)=\frac{1}{2^n} and, for m\leq n, d(m,n)=\sum_{k=m}^{n-1} d(k,k+1). (This example corresponds to considering the subset \{\frac{1}{2^n}|n\geq -1\} with the usual topology of \mathbb{R}). Since B(n,\frac{1}{2^n})=\{n\} (open ball), this defines the discrete topology. Yet you can see that the sequence (n)_{n\geq 0} is a Cauchy sequence that doesn't converge.
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    Isn't completeness a property of metrics rather than of topologies?

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

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

    Now \mathbb R is d-complete but not \bar d-complete since (n)_{n\in\mathbb N} is a \bar d-Cauchy sequence that does not converge to anything in (\mathbb R,\bar d).
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    Quote Originally Posted by halbard View Post
    Isn't completeness a property of metrics rather than of topologies?
    Sure, with the exception of compact spaces. Compacity is a topological property, yet it implies completeness for any compatible metric (if there is one). The question is to prove that compacity is the only exception. This is an interesting problem, I think.
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    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?
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  9. #9
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    Quote Originally Posted by Laurent View Post
    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 \mathbb{N} defined by d(n,n)=0, d(n,n+1)=\frac{1}{2^n} and, for m\leq n, d(m,n)=\sum_{k=m}^{n-1} d(k,k+1). (This example corresponds to considering the subset \{\frac{1}{2^n}|n\geq -1\} with the usual topology of \mathbb{R}). Since B(n,\frac{1}{2^n})=\{n\} (open ball), this defines the discrete topology. Yet you can see that the sequence (n)_{n\geq 0} is a Cauchy sequence that doesn't converge.
    ...Ouch.

    Well, I was thinking of using something like Alexandroff's compactification by a point. Like when you identify \mathbb{R} ^n with A:= \mathbb{S} ^n - \{ e_n \} where \mathbb{S} ^n := \{x \in \mathbb{R} ^{n+1} : \Vert x \Vert =1 \} and any sequence converging to \infty (or e_n) is a Cauchy sequence in A that doesn't converge in \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?
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    Quote Originally Posted by Jose27 View Post
    Well, I was thinking of using something like Alexandroff's compactification by a point. Like when you identify \mathbb{R} ^n with A:= \mathbb{S} ^n - \{ e_n \} where \mathbb{S} ^n := \{x \in \mathbb{R} ^{n+1} : \Vert x \Vert =1 \} and any sequence converging to \infty (or e_n) is a Cauchy sequence in A that doesn't converge in \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?
    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.
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    Quote Originally Posted by Opalg View Post
    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.
    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.
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    Quote Originally Posted by Laurent View Post
    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.
    That's true, that idea will only work for a locally compact space.

    Quote Originally Posted by Laurent View Post
    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.
    The Stone–Cech compactification (of a noncompact space) is enormous, and as far as I know it is never metrisable.
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    Okay, but the topology we induce on X \bigcup \{ p \} is not necessarily that of the ususal compactification: For example when we do the stereographic projection of \mathbb{S} ^n on \mathbb{R} ^n the topology we give to \mathbb{S} ^n is \tau= \{ U \subset X:U is open in X \} \bigcup \{ p \in U : U=(X \bigcup \{ p \} )-V where V \subset X is compact \} and X= \mathbb{R} ^n.
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  14. #14
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    Two metrics d_1 and d_2 on same X are equivalent, if they produces same topologies, i.e. {\tau}_{d_1}={\tau}_{d_2}.
    For general case, if the staments is true, remind that if (X,d) isn't compact, he contain sequence {\{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.
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