Is there any metric $d$ of $\mathbb R^n$, $n<\infty$ such that $\mathbb R^n$ is bicompact and no norm induces $d$

Question

There are some simple metrics can't yielded by norm .But add bicompact,I can't structure such example.

In fact ,I want to know the condition of metric can be yielded by norm.

Sorry for my poor English.

Answer 1

If a metric $d$ on $\mathbb R^n$ is induced by a norm, then the metric space $(\mathbb R^n,d)$ is unbounded and therefore cannot be compact (I interpret bicompact as an old-fashioned term for compact). Indeed, take any nonzero vector $v$ and positive number $t$: then $$ d(tv,0)=\|tv\|=t\|v\| $$ and this tends to infinity as $t\to\infty$.

Thus, any metric with respect to which $\mathbb R^n$ is compact is not induced by a norm.

Also, any such metric must be incompatible with the standard topology on $\mathbb R^n$, since compactness is a topological invariant. This makes it hard to come up with any natural examples.

But it's easy to come up with non-natural examples. There is a bijection $f:\mathbb R^n\to [0,1]$, since both sets have cardinality continuum. Then define $$ d(x,y) = |f(x)-f(y)|,\quad x,y\in\mathbb R^n $$ The metric space $(\mathbb R^n,d)$ is compact, since the correspondence $x\mapsto f(x)$ is now an isometry.