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Triangle inequality metric space

  1. Jun 3, 2010 #1
    1. The problem statement, all variables and given/known data

    Let [itex](X,\theta)[/itex] be a metric space. Take [itex] K > 0 [/itex]and define.

    [itex]\theta : X \cross X \rightarrow \real_{0}^{+}[/itex], [itex](x,y)\rightarrow \frac{K\phi(x,y)}{1+K\phi(x,y)}[/itex]

    Show that [itex](X,\theta)[/itex] is a metric space.


    2. Relevant equations

    can someone please check my triangle inequality?

    3. The attempt at a solution


    [itex]\phi(x,z) \leq \frac{K\phi(x,y)}{1+K\phi(x,y)}[/itex]

    [itex]\leq \mid \frac{K\phi(x,y)}{1+K\phi(x,y)}\mid + \mid \frac{K\phi(y,z)}{1+K\phi(y,z)}\mid[/itex]

    [itex]= \mid \frac{K\phi(x,y)}{1+K\phi(x,y)} + \frac{K\phi(y,z)}{1+K\phi(y,z)}\mid[/itex]

    [itex]=\phi(x,y)+\phi(y,z)[/itex]
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 3, 2010 #2

    lanedance

    User Avatar
    Homework Helper

    apologies if I'm missing something.. but what is [itex]\phi(x,y)[/itex]?

    and shouldn't you be trying to show the triangle inequality holds for [itex]\theta(x,y)[/itex]?
     
  4. Jun 3, 2010 #3
    Yeah your right it is a typo the metric space should be [itex](X,\phi)[/itex]


    so triangle inequality is

    [itex]\phi(x,y) \leq \frac{K\phi(x,)}{1+K\phi(x,y)}[/itex]
    [itex]\leq \mid \frac{K\phi(x,z)}{1+K\phi(x,z)}\mid + \mid \frac{K\phi(z,y)}{1+K\phi(z,y)}\mid[/itex]

    [itex]= \mid \frac{K\phi(x,z)}{1+K\phi(x,z)} + \frac{K\phi(z,y)}{1+K\phi(z,y)}\mid[/itex]

    [itex]=\phi(x,z)+\phi(z,y)[/itex]
     
  5. Jun 4, 2010 #4
    Is it ok now?
     
  6. Jun 4, 2010 #5

    Mark44

    Staff: Mentor

    To show that [itex](X, \phi)[/itex] is a metric space, you need to show:
    1) [itex]\phi(x, y) = 0~\text{iff}~x = y[/itex]
    2) [itex]\phi(x, y) = \phi(y, x)[/itex]
    3) [itex]\phi(x, y) + \phi(y, z) \geq \phi(x, z)[/itex]

    Presumably you have already shown 1 and 2 and are working on 3 here (with some corrections).

    [tex]\phi(x,y) = \frac{K\phi(x, y)}{1+K\phi(x, y)}[/tex]
    In the line above, it should be =, by how phi(x, y) is defined in post #1, although how it is defined is hazy, since phi(x, y) is defined in terms of itself.

    How do you get to the next line? What's the justification here?
    [tex]\leq \mid \frac{K\phi(x, z)}{1+K\phi(x, z)}\mid + \mid \frac{K\phi(z, y)}{1+K\phi(z, y)}\mid[/tex]
     
  7. Jun 4, 2010 #6
    I hope this clears it up. Sorry I'm not to good at latex.


    Let[itex](X,\phi)[/itex]be a metric space. Take [itex] K > 0 [/itex]and define.
    [itex]\theta : X \cross X \rightarrow \real_{0}^{+}[/itex],[itex](x,y)\rightarrow\frac{K\phi(x,y)}{1+K\phi(x,y)}[/itex]




    show that [itex](X,\theta)[/itex]is a metric

    so the triangle inequality
    [itex]\theta(x,y) = \frac{K\phi(x,)}{1+K\phi(x,y)}[/itex]
    [itex]\leq \mid \frac{K\phi(x,z)}{1+K\phi(x,z)}\mid + \mid

    \frac{K\phi(z,y)}{1+K\phi(z,y)}\mid[/itex]
    [itex]= \mid \frac{K\phi(x,z)}{1+K\phi(x,z)} +

    \frac{K\phi(z,y)}{1+K\phi(z,y)}\mid[/itex]

    [itex]=\phi(x,z)+\phi(z,y)[/itex]

    My justification is because I was given that [itex](X,\phi)[/itex] was a metric

    space, we know that [itex]\phi(x,y)[/itex] is non negative. That in turn with [itex]

    K > 0 [/itex] ensures that [itex]\theta : X \cross X \rightarrow

    \real_{0}^{+}[/itex] is a well defined non negative function.
     
    Last edited: Jun 4, 2010
  8. Jun 4, 2010 #7
    It suppose to be
    show
    [itex](X,\theta)[/itex]
    is a mertic
     
  9. Jun 4, 2010 #8

    Mark44

    Staff: Mentor

    OK, I think I understand what you're trying to say, now. You are given that [itex](X, \phi)[/itex] is a metric space (which means that [itex]\phi[/itex] is a metric).

    There is another function [itex]\theta[/itex], where [itex]\theta[/itex]: X x X --> [0, [itex]\infty[/itex]), with K > 0 and
    [tex]\theta(x, y) = \frac{K\phi(x, y)}{1 + K\phi(x, y)}[/tex]

    You are trying to show that [itex]\theta[/itex] is a metric (not mertic), which is the same as saying that [itex](X, \theta)[/itex] is a metric space. (A metric space is a set together with a function that measures distance between elements of the set.) Part of the definition of a metric is that it is nonnegative, so you don't need the absolute values. The other parts of the definition I showed in post 5.

    Since [itex]\phi[/itex] is a metric, it satisfies the triangle inequality. I see that you are trying to work this into your proof, but what you have doesn't look right to me. Certainly you can say that [itex]\phi(x, y) \leq \phi(x, z) + \phi(z, y)[/itex], but you also have [itex]\phi(x, y)[/itex] in the denominator, and you can't just substitute things in directly. Getting this right will require some more thought.
     
  10. Jun 4, 2010 #9

    lanedance

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    Homework Helper

    i think Mark's onto it now, but i have to say the fact that there are 2 different posters & theta and phi have been interchanged (incorrectly as i read it now) in every post has made it pretty confusing...
     
  11. Jun 4, 2010 #10

    Mark44

    Staff: Mentor

    Amen to that, lanedance!
     
  12. Jun 4, 2010 #11
    Sorry I'm getting lost here. Do I need to somehow get [itex]\phi(x, y)[/itex] out of the denominator?
     
  13. Jun 4, 2010 #12

    Mark44

    Staff: Mentor

    You can use the fact that [itex]\phi[/itex] is a metric to say this:
    [tex]K\phi(x, y) \leq K\phi(x, z) + K\phi(z, y)[/tex]

    because [itex]\phi[/itex] satisfies the triangle inequality.

    But you can't just say that
    [tex]\frac{K\phi(x, y)}{1+K\phi(x, y)} \leq \frac{K\phi(x, z)}{1+K\phi(x, z)} + \frac{K\phi(z, y)}{1+K\phi(z, y)}[/tex]

    without some justification, particular regarding those denominators.
     
  14. Jun 5, 2010 #13
    But isn't the justification that if K >0 and
    [tex]K\phi(x, y) [/tex] is a metric thats non negative

    That it can't be > [tex] \frac{K\phi(x, z)}{1+K\phi(x, z)} + \frac{K\phi(z, y)}{1+K\phi(z, y)}[/tex]?

    You've got that the denominator = at least 1. So if x = z the function is 0.
     
  15. Jun 5, 2010 #14

    Mark44

    Staff: Mentor

    No, there's more to it than that. And it's not [itex]K\phi(x, y) [/itex] that is a metric - [itex]\phi [/itex] is a metric, so [itex]\phi(x, y) \geq 0[/itex] for any x and y in X.

    It's easy enough to show that [itex]\theta(x, y) \geq 0 [/itex] because of how [itex]\theta(x, y)[/itex] is defined.

    Have you already proved that [itex]\theta(x, y) = 0[/itex] iff x = y? Have you already proved that [itex]\theta(x, y) = \theta(y, x)[/itex]?

    If so, then you still need to prove that [itex]\theta(x, y) \leq \theta(x, z) + \theta(z, y)[/itex]. The expression you have in post #13 is equal to [itex]\theta(x, z) + \theta(z, y)[/itex]. You need to show that this value is >= [itex]\theta(x, y)[/itex].
     
  16. Jun 5, 2010 #15
    @boneill3: Can you explain in more detail what is confusing you? It took some work, but I've convinced myself that θ is a metric.
     
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