Ultrametric on a normed space (real or complex)

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 $$||x+y|| \leq ||x||+||y||$$

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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...

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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.

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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