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Math Help - Ultrametrics

  1. #1
    Super Member Aryth's Avatar
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    Ultrametrics

    I've been asked to show that a particular metric is actually an ultrametric. I'm a little lost, and I'll post what I have.

    Here's the metric we're supposed to look at:

    Let X be any nonempty set. Define a metric \mu : X \times X \to \mathbb{R} by letting \mu (x,x) = 0 and \mu (x,y) = \mu (y,z) be any number in the interval [1,2] when x \neq y.

    And the problem is to show that this is an ultrametric, or that \mu (x,z) \leq Max\[\mu (x,y), \mu (y,z)\] for all x,y,z \in X.

    Here's my attempt at the proof:

    First, since \mu is a metric, we are given that \mu (x,z) \leq \mu (x,y) + \mu (y,z). If x = z, then \mu (x,z) = 0 \leq Max\[\mu (x,y), \mu (y,z)\]. If x \neq z and x = y OR y = z, then

    \mu (x,z) \leq \mu (x,y) + \mu (y,z) = \mu (x,y) + 0 = Max\[\mu (x,y),\mu (y,z)\]
    or
    \mu (x,z) \leq \mu (x,y) + \mu (y,z) = 0 + \mu (y,z) = Max\[\mu (x,y), \mu (y,z)\]
    The only remaining case is when x\neq y\neq z.

    I can't figure out the solution for this. It may be that I am taking the wrong approach. Any help is greatly appreciated.
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  2. #2
    Super Member Aryth's Avatar
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    Re: Ultrametrics

    Nothing?
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    Re: Ultrametrics

    If x\neq y\neq z then by the definition of the metric \mu(y,x) = \mu(x,z) = \mu(z,y) \in [1,2]. Hence \mu(x,y) = \mu(y,z) = \mu(x,z) = \max[\mu(x,y),\mu(y,z)].
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    Super Member Aryth's Avatar
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    Re: Ultrametrics

    Quote Originally Posted by SlipEternal View Post
    If x\neq y\neq z then by the definition of the metric \mu(y,x) = \mu(x,z) = \mu(z,y) \in [1,2]. Hence \mu(x,y) = \mu(y,z) = \mu(x,z) = \max[\mu(x,y),\mu(y,z)].
    I really appreciate your response. I'm just wondering how the definition of the metric implies that they are all equal... I looked at all of the properties of the metric that are in the definition I was given, and none of them outright imply this. The closest I can see is the property that \mu (x,y) = \mu (y,x) for all x,y \in X, but that doesn't lead me straight to where you were.
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    Re: Ultrametrics

    That definition gives: \mu(y,x) = \mu(x,z) since y\neq x. \mu(x,z) = \mu(z,y) since x \neq z. \mu(z,y) = \mu(y,x) since z \neq y.

    Also, \mu(y,x) = \mu(x,y) and \mu(z,y) = \mu(y,z) because \mu is a metric (so it is symmetric).

    Put that all together, and you get that everything is equal.
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    Re: Ultrametrics

    Quote Originally Posted by Aryth View Post
    Let X be any nonempty set. Define a metric \mu : X \times X \to \mathbb{R} by letting \mu (x,x) = 0 and \mu (x,y) = \mu (y,z) be any number in the interval [1,2] when x \neq y.
    The bolded text is where I got those equalities, by the way.
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    Re: Ultrametrics

    Also, if that is not the case \left( \mbox{if that was a typo, and it should have read }\mu(x,y) = \mu(y,x)\mbox{ instead of }\mu(x,y) = \mu(y,z)\right), then here is an example of a set X where \mu is not an ultrametric. Example:

    X = \left\{(0,0),(1,0),\left(0,\dfrac{5}{4}\right) \right\} \subset \mathbb{R}^2

    Let \mu be the standard Euclidean metric, x = (1,0), y = (0,0), z = \left(0,\dfrac{5}{4}\right).
    \mu(x,y) = 1\in [1,2], \mu(y,z) = \dfrac{5}{4}\in [1,2] and \mu(x,z) = \dfrac{3}{2}\in [1,2]. Hence \mu(x,z) > \max[\mu(x,y),\mu(y,z)].
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  8. #8
    Super Member Aryth's Avatar
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    Re: Ultrametrics

    I see now. I completely overlooked that fact. Thanks a lot for your help!
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