Thread: 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 be any nonempty set. Define a metric by letting and be any number in the interval when .

And the problem is to show that this is an ultrametric, or that for all .

Here's my attempt at the proof:

First, since is a metric, we are given that . If , then . If and OR , then

or

The only remaining case is when .

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?

If then by the definition of the metric . Hence .

Originally Posted by SlipEternal

If then by the definition of the metric . Hence .

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 for all , but that doesn't lead me straight to where you were.

That definition gives: since . since . since .

Also, and because is a metric (so it is symmetric).

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

Originally Posted by Aryth

Let be any nonempty set. Define a metric by letting and be any number in the interval when .

The bolded text is where I got those equalities, by the way.

Also, if that is not the case , then here is an example of a set where is not an ultrametric. Example:



Let be the standard Euclidean metric, .
, and . Hence .

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