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Which metric spaces are isometric to $(\mathbb{R}^n, d_E)$?

Two metric spaces $(M_1, d_1)$ and $(M_2, d_2)$ are called isometric, when an isometry between them exists. An isometry is an isomorphism $\varphi: M_1 \rightarrow M_2$ such that

$$\forall p_1, p_2 \in M_1: d_1(p_1, p_2) = d_2(\varphi(p_1), \varphi(p_2))$$

is true.

Martin Thoma
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  • I think it's only isometric to itself. Isn't this an old Banach result? – Henno Brandsma Feb 10 '15 at 14:21
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    I wouldnt think so. Multiplying the euclidean metric with a constant C > 0 yields a different metric, but you could easily find an isometry for this case. – user159517 Feb 10 '15 at 14:38
  • @moose in your definition of isometry, I think you want $f$ to be a bijection, or else there could be some alien points in $M$ that do not correspond to any points in $\mathbb R^n$. – fonini Feb 10 '15 at 14:47
  • What kind of answer are you looking for besides "the metric spaces which are isometric to $(\mathbb{R}^n, d_E)$"? – Najib Idrissi Feb 10 '15 at 15:05
  • @NajibIdrissi E.g. "All isometric spaces are only $(R^n, c \cdot d_E)$, where $c>0$" is a constant or some more exotic metrics which are isometric. Or some other connection. A criterion which I can use to tell if a metric is "basically" only the Euclidean metric. These kinds of answers. – Martin Thoma Feb 10 '15 at 15:11

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