Ultrametrics

**Aryth** · Oct 15th 2013, 01:29 PM

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.

**Aryth** · Oct 17th 2013, 11:48 AM

Nothing?

**SlipEternal** · Oct 17th 2013, 12:31 PM

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 $\mu(x,y) = \mu(y,z) = \mu(x,z) = \max[\mu(x,y),\mu(y,z)]$ .

**Aryth** · Oct 17th 2013, 01:09 PM

Originally Posted by SlipEternal

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.

**SlipEternal** · Oct 17th 2013, 01:21 PM

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.

**SlipEternal** · Oct 17th 2013, 01:23 PM

Originally Posted by Aryth

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.

**SlipEternal** · Oct 17th 2013, 02:09 PM

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 $\mu(x,z) > \max[\mu(x,y),\mu(y,z)]$ .

**Aryth** · Oct 17th 2013, 08:59 PM

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

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