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I need help regarding the following two exercises:

a) Show that $(\mathbb R^2, d_2)$ and $(\mathbb R,d_1)$ $d_2,d_1$ being the respective euclidean metrics, are not isometric isomorphic, i.e. there is no bijective isometric map between those metric spaces.
b) Let d be the restriction of the euclidean metric of $\mathbb R$ on $(-1,1) \times (-1,1)$. Find a metric $d':\mathbb R \times \mathbb R\to \mathbb R$ such that the spaces $((-1,1),d)$ and $(\mathbb R,d')$ are isometric isomorphic.

For a) I think I have solution but I am not really sure:
Does it suffice to say that for every possible candidate $f$ for an isomorphism it can't be injective since it must be norm preserving, i.e. $$||f(x)||_2 = ||x||_2, \quad ||f(-x)||_2 = ||-x||_2 = ||x||_2$$ with $||\cdot||_2$ being the respetive euclidean norm?

For b) I don't really know how to start.

lasik43
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  • $\mathbb{R}^2$ and $\mathbb{R}$ are not isomorphic as vector spaces over $\mathbb{R}$ (see here), so there cannot be an isometric isomorphism. For b), picture that $(-1,1)\times (-1,1)$ can be "stretched" infinitely long towards $\mathbb{R}^2$. Describe this stretching as a bijective function $f \colon (-1,1)\times (-1,1) \to \mathbb{R}^2$, then I am pretty sure that $d'= d \circ f^{-1}$ will provide the desired metric. – lattice Jun 08 '16 at 13:26
  • So is my argumentation false? – lasik43 Jun 08 '16 at 13:29
  • Well, only $||f(x)||_2=||x||_2=||f(-x)||_2$ is not enough, since for any norm value $x>0$ there are also always two numbers in $\mathbb{R}$ which take on this value, i.e. $x$ and $-x$. But if you extend your argumentation (or maybe you already meant that and I got it wrong), saying that: For $x>0$ there are two numbers in $\mathbb{R}$ with norm $x$, but in $\mathbb{R}^2$ there are infinitely many such numbers (all points of a circle with the radius $x$), then it would be correct, I guess. But anyway, how should there be an isometric isomorphism if there is not even an isomorphism? – lattice Jun 08 '16 at 13:38

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