Are all the finite dimensional vector spaces with a metric isometric to $\mathbb R^n$?
My goal is to claim that in any finite dimensional vector space, equipped with a metric, a closed-bounded subset is compact.
I know that all finite dimensional inner-product spaces are equivalent, but I never heard it about metric so my hunch is that I'm wrong. Yet I'd love to know for sure.
Is it right for normed space though?
thanks!
Best Answer
If $V$ and $W$ are normed vector spaces of the same finite dimension $n$, then $V$ and $W$ are isomorphic as topological vector spaces, i.e., there is a a vector space isomorphism $f : V \rightarrow W$ such that both $f$ and $f^{-1}$ are continuous. However, if $n \ge 2$, $V$ and $W$ may not be isometrically equivalent, i.e., it may be impossible to choose such an $f$ that preserves distances.
To see that $V$ and $W$ are isomorphic as topological vector spaces, pick a basis for each of $V$ and $W$ and use these to define a vector space isomorphism $f : V \rightarrow W$. Then argue that if $B_V$ is the unit ball in $V$ and $B_W$ is the unit ball in $W$, then $f[\lambda B_V] \subseteq B_W \subseteq f[\mu B_V]$ for some positive $\lambda, \mu \in \mathbb{R}$, and conclude that $f$ and $f^{-1}$ are both continuous.
To see that $V$ and $W$ need not be isometrically equivalent, take $V$ to be $\mathbb{R}^2$ with the euclidean norm, $\|(x, y)\|_V = \sqrt{x^2+y^2}$, and take $W$ to be $\mathbb{R}^2$ with the sup-norm, $\|(x, y)\|_W = \mathsf{sup}(|x|, |y|)$. Let $S_V$ and $S_W$ be the unit circles in $V$ and $W$ respectively. (So $S_V$ is the euclidean circle of unit radius centred at the origin and $S_W$ is the square with vertices $(\pm 1, \pm1)$.) Then for any $v_1 \in S_V$ there is a unique $v_2 \in S_V$ (namely $-v_1$) such that $\|v_1 - v_2\|_V = 2$, and this property must clearly be preserved in any isometric image of $V$. However, this property fails in $W$: with $w_1 = (-1, 1)$, $\|w_1 - w_2\|_W = 2$ for any $w_2$ on the line segment from $(1, 1)$ to $(1, -1)$ and all such $w_2$ lie on $S_W$. So $W$ cannot be an isometric image of $V$.