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I want to find a metric $d'$ sucht that $(\mathbb{R}, d')$ and $(X, d)$ with $X = (-1,1)$ and $d(x,y) = |x-y|$ are isometric. I tried the homeomorphisms $$ f:\mathbb{R} \to (-1,1) ~ \textrm{ with } ~ x \mapsto \frac{1}{x+1} + \frac{1}{x-1} $$ and $$ g:(-1,1) \to \mathbb{R} ~ \textrm{ with } ~ x \mapsto \tan\left(\frac{\pi}{2} x\right) $$ but i have no idea how to find the metric $d'$?

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Let $\phi\colon \mathbb R \to (-1,1)$ a homeomorphism. Define $d'(x,y) := \left|\phi(x) - \phi(y)\right|$. Then $d$ is as wished and $\phi$ is an isometry. –  martini Oct 16 '12 at 10:20

2 Answers 2

up vote 3 down vote accepted

Once you pick your favorite homeomorphism $h:\Bbb R\to (-1,1)$, you can define $$d'(a,b):=d(h(a),h(b))$$ And this is applicable to your functions by $h:=f$ or $h:=g^{-1}$.

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If you have a injective $f : \mathbb{R} \to X$, you can use that to define a metric $d'$ on $\mathbb{R}$ which makes $f$ isometric. Just set $$ d'(x,y) = |f(x) - f(y)| $$

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