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Equivalent Metrics From Clopen Sets

  1. Apr 5, 2015 #1
    1. The problem statement, all variables and given/known data


    Prove that if ##(X,d)## is a metric space and ##C## and ##X \setminus C## are nonempty clopen sets, then there is an equivalent metric ##\rho## on ##X## such that ##\forall a \in C, \quad \forall b \in X \setminus C, \quad \rho(a,b) \geq 1##.

    I know the term "clopen" is not a very formal definition, at least not to my knowledge, but it does describe the two properties of the given sets: they are both open and closed.



    2. Relevant equations


    3. The attempt at a solution

    Would I have to show that the metric ##\rho## satisfies the properties of a metric or would I need to show that the metric ##d## and the metric ##\rho## generate the same topology to show they are equivalent?

    If I define a function ##\rho: X \times X \to \mathbb{R}_+## by the following:

    if both ##x,y## lie in either ##C## or ##X \setminus C##, then ##\rho(x,y) = d(x,y)##. Otherwise, let ##\rho(x,y) = d(x,y) +100##, say. How do I proceed to show that this function satisfies the properties of a metric, that it is equivalent to ##d##, and that it satisfies the desired property of ##\rho(x,y) \geq 1 ##?

    As a side thought, would using sequences be useful here, or would this be a completely different approach?

    Any help with this problem is greatly appreciated as I do not know where to begin or how to proceed in writing up a correct proof.

    Thanks in advance for your time and patience.
     
  2. jcsd
  3. Apr 5, 2015 #2

    WWGD

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    Do you know what property follows , i.e., is equivalent to having a clopen set other than X and the empty set?
     
  4. Apr 5, 2015 #3
    Would it have something to do with being connected? My textbook states that a space ##X## is connected if there do not exist open subsets ##A, B## of ##X## such that ##A \neq \emptyset, \quad B \neq \emptyset, \quad A \cap B = \emptyset, \quad A \cup B = X##. Is this equivalent? Thank you for your prompt response.
     
  5. Apr 5, 2015 #4

    WWGD

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    Precisely, good going.
     
  6. Apr 5, 2015 #5
    Ok, so if ##C \cup X \setminus C = X##, then this creates a separation, according to what I am reading from Fred H. Croom's Principles of Topology. How do I tie this together with metric spaces to show equivalence? Again, many thanks for your feedback.
     
  7. Apr 5, 2015 #6

    Dick

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    There are two notions of equivalent metrics. http://en.wikipedia.org/wiki/Equivalence_of_metrics Which one are you using? I'm guessing you are just looking for topological equivalence.
     
  8. Apr 5, 2015 #7
    Yes, we are using topological equivalence. To be honest, I do not think I have seen strong equivalence before. Thank you for this clarification.
     
  9. Apr 5, 2015 #8

    Dick

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    Well then, just go ahead with your idea. Show ##\rho## is a metric. That should be easy. Then show they both generate the same topology. That's where you use the 'clopen'.
     
  10. Apr 5, 2015 #9

    WWGD

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    The basic properties of [itex]\rho [/itex] as a metric follow from those of [itex]d(x,y) [/itex]. Just consider the cases for x,y,z are not all in the same component.
     
  11. Apr 6, 2015 #10
    So by construction, ##\rho(x,y) = d(x,y)## if both ##x,y \in C## or ##x,y \in X \setminus C##. As for considering the cases where ##x,y,z## are not all in the same component, my book defines component of a topological space as a connected subset ##C## of ##X## which is not a proper subset of any connected subset of ##X##. There is also a list of properties of the components which follow the definition, but I am unsure how to properly use this.

    I get the feeling that showing the triangle inequality would look something like this:

    if ##\rho(x,z) \leq \rho(x,y) + \rho(y,z)##, then let ##\rho(x,y) = d(x,y)+50## and let ## \rho(y,z) = d(x,y)+50## so that we get ##\rho(x,z) \leq d(x,y) +100##. Does this seem correct? Many thanks for your time and assistance.
     
    Last edited: Apr 6, 2015
  12. Apr 6, 2015 #11

    Dick

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    WWGD was just using the word 'component' to refer to either ##C## or ##X \setminus C##. You know ##d## is a metric so use that to show the ##\rho## you defined in first post is also a metric.

    So you can start by assuming ##d(x,y) \le d(x,z) + d(z,y)## for any ##x, y, z##. You want to show ##\rho(x,y) \le \rho(x,z) + \rho(z,y)##. Just consider the cases. It's pretty obvious if ##x##, ##y## and ##z## lie in the same set, yes? Now just consider other cases, like ##x## and ##y## lie in the same set and ##z## lies in the other. Or ##x## and ##z## are in the same set and ##y## is in the other. Is the ##\rho## inequality also true then? Just substitute for what ##\rho## is in terms of ##d##. Do you still get true inequalities?
     
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