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?


  • $\begingroup$ Try with the discrete metric. More hypotheses on the metric are necessary. $\endgroup$ – egreg Nov 21 '14 at 21:45
  • $\begingroup$ metric-isomorphic is the same as isometric? $\endgroup$ – janmarqz Nov 21 '14 at 22:09
  • $\begingroup$ @janmarqz: ya, that's what i ment $\endgroup$ – user188400 Nov 21 '14 at 22:16
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    $\begingroup$ When you ask questions like these, always make sure all the different properties are tied together. You should have easily spotted the problem from the fact that you're talking about "finite dimensional vector spaces with a metric" - but a metric doesn't care about the vetor space structure, so the requirement that your set be a vector space is immediately forgotten. $\endgroup$ – Jack M Nov 21 '14 at 22:24
  • $\begingroup$ mm @JackM: vector space can be relevant. especially if you consider that normed spaces are vector spaces, so it might be $\endgroup$ – user188400 Nov 21 '14 at 22:26

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 $V$.


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