norms on a field and induced metric

Question

Let $F$ be a field. A norm on $F$ is a map $|\,|:F\rightarrow\mathbb{R}$ with conditions:

• $|x|\ge 0$ and $|x|=0$ if and only if $x=0$.

• $|xy|=|x|.|y|$

• $|x+y|\le |x|+|y|$

A norm $|\,|$ is said to be non-archimedean if

• $|x+y|\le \max \{ |x|,|y|\}$ for all $x,y\in F$.

Observations:

(1) Let $|\, |$ be a norm on $F$. This defines a metric on $F$ by $d(x,y)=|x-y|$.

(2) If $|\,|$ is non-archimedean then the metric $d$ is bounded: there is $M\in\mathbb{R}$ such that $|x|\le M$ for all $x\in F$ or equivalently $d(x,y)\le M$ for all $x,y\in F$.

Q.1 Are there some bounded metrics on $F$ which is not induced by a non-archimedean norm on $F$?

Q.2 Are there metrics on $F$ which are not induced by a norm on $F$?

For Q.2 I was thinking to take usual metric on $\mathbb{R}$, and compose with a homeomprphism $\mathbb{R}\rightarrow \mathbb{R}$, $x\mapsto 2x$ to get a new metric on $\mathbb{R}$; this is not induced by any norm on $\mathbb{R}$ since for norm, $|1|=1$.

Q.1 I am not getting some obvious answers as in Q.2.

Answer 1

A positive answer to Q.2 is obviously also a positive answer to Q.1, so you probably want to rephrase Q.1 as "Are there some bounded metrics on $F$ which are induced by norm, but are not induced by a non-archimedean norm"?

As for question 2, consider the metric $$d((x_1,y_1),(x_2,y_2))=(|x_1-x_2|^p+|y_1-y_2|^p)^{1/p}$$ defined on the complex field $\mathbb{C}$, where a complex number is represented by $x+iy$. If $0<p<1$, the unit ball of this metric (around zero, say) is not convex. It is easy to check that the unit ball of a norm must be convex, hence this class of metrics are not induced by a norm.

Answer 2

For any metric space, there is a bounded metric inducing the same topology on it. For example, one can replace a metric $d$ by $d'(x,y)=\min(d(x,y),1)$. So on $\Bbb R$ one has the bounded metric $\min(|x-y|,1)$. This induces the standard topology, not the totally disconnected topology induced by a non-Archimedean topology.