1. ## 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.

Nothing?

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

4. ## Re: Ultrametrics

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.

5. ## 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.

6. ## Re: Ultrametrics

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.

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

8. ## Re: Ultrametrics

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