Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given some normed space $E$ (real or complex), why is it impossible that $E$ can't be an ultrametric space?

My professor briefly said something along these lines today and I didn't follow...

Also, I am not sure what the strong triangle inequality would be for a normed space... Would it be $\forall x,y \in E$ we have $$||x+y|| \leq ||x||+||y||$$

?

Or does it include 3 elements of $E$ like the definition of the ultrametric $\forall x,y,z \in E$ we have $$d(x,z) \leq d(x,y)+d(y,z)$$

I feel like I am missing something here and I don't know where to start... Any pointers? I am not as familiar with normed spaces...

share|improve this question

1 Answer 1

up vote 3 down vote accepted

The strong triangle inequality actually says that for any $x,y,z\in E$,

$$\|x-y\|\le\max\{\|x-z\|,\|y-z\|\}\;;$$

however, this is equivalent to saying that for any $x,y\in E$,

$$\|x+y\|\le\max\{\|x\|,\|y\|\}\;.\tag{1}$$

In fact this can be strengthened: one can prove that if $(1)$ holds, then $\|x+y\|=\max\{\|x\|,\|y\|\}$ whenever $\|x\|\ne\|y\|$. (The easy proof is in Wikipedia.)

Open balls in an ultrametric space are automatically also closed, so every ultrametric space has a clopen base and is therefore zero-dimensional. In particular, it’s totally disconnected. If $E$ is a real or complex normed space, then for any non-zero $x\in E$ the set $\{\lambda x:1\le\lambda\le 2\}$ is connected, since it’s a continuous image of $[0,1]$, so $E$ cannot be totally disconnected. Thus, the norm metric on $E$ cannot be an ultrametric.

share|improve this answer
    
That definition of the ultrametric makes a lot more sense than what I thought. So because in any subset of an ultrametric space, there is a clopen ball, then that subset is disconnected, so $E$ is totally disconnected. I don't really see how that set is connected (connected by path?) but I will trust what you are saying. Thanks a lot! –  Craig Wilson Oct 31 '13 at 1:06
    
@Danielle: It’s a general theorem: if $f:X\to Y$ is continuous, and $C$ is a connected subset of $X$, then $f[C]$ is a connected subset of $Y$. (In the special case in which $X$ and $Y$ are $\Bbb R$ this is basically just the intermediate value theorem.) You’re welcome! –  Brian M. Scott Oct 31 '13 at 1:13

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.