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Is every metric on the real line translation invariant?

i.e. If $(\mathbb R,d)$ is any metric space then is it true in general that

$d(a + x, b + x) = d(a,b)$ for all $a,b,x \in \mathbb R.$

Please help me in understanding this fact.

Thank you in advance.

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No. Take for example $d(x,y)= |\arctan (x)-\arctan (y)|$. Show that this is a distance on the real line. Is it invariant by translation?

P.S. Note that $d(x,y)=|f(x)-f(y)|$ is a distance on the real line as soon as $f:\mathbb{R}\to\mathbb{R}$ is injective.

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No. A visual example would be to imagine the line $\Bbb R$ embedded into $\Bbb R^2$ or $\Bbb R^3$ in some curvy way, e.g. as the graph of $\sin(x)$, etc. Write $\iota:\Bbb R\rightarrow \Bbb R^n$ for such an embedding, e.g. $\iota(x)=(x,\sin(x))$.

If you now view the distance between points of the line with respect to this higher dimensional space, you see that it depends on their actual position in space and not just their distance in the 1D-coordinate on the line. Formally, take the metric $d(a,b):=\|\iota(a)-\iota(b)\|$, which is far from translation invariant in the general case.

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Generalization of the answer of Robert Z: Let be $f:\Bbb R\longrightarrow(X,\delta)$ injective, $(X,\delta)$ metric space. The function $$d(x,y) = \delta(f(x),f(y))$$ will be a metric but almost never will be translation-invariant.

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