Let $(X,d_1)$ and $(X,d_2)$ be two metric spaces on the same set $X$. Is there any relation between $d_1$ and $d_2$ being equivalent and $(X,d_1)$ and $(X,d_2)$ being isometric? If not, can anyone give examples where $d_1$ and $d_2$ are equivalent but $(X,d_1)$ and $(X,d_2)$ are not isometric; and where $d_1$ and $d_2$ are not equivalent, but $(X,d_1)$ and $(X,d_2)$ are isometric?
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1in which sense precisely do you mean that the two metrics are equivalent? – Ittay Weiss Jun 13 '13 at 09:07
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In the sense that they induce exactly the same topology on $X$; or, id from $X$ to $X$ is a homeomorphism. – Mr. Chip Jun 13 '13 at 09:13
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1Read this link to get further insight – Srijan Jun 13 '13 at 10:04
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@ittay weiss; what if I defined equivalent metrics to be; Two metrics $d_1$ and $d_2$ are said to be equivalent if there exists two positive real numbers $\alpha$ and $\beta$ s.t $\alpha d_1(x,y)≤d_2(x,y)≤\beta d_1(x,y)$ ?? – Ibrahim Islam Oct 02 '21 at 18:14
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If I understand your question correctly, there is no relation between being equivalent and being isometric.
To see that "equivalent" does not imply "isometric", take $X=\mathbb R$ with $d_1(x,y)=\vert x-y\vert$ and $d_2(x,y)=\frac{\vert x-y\vert}{1+\vert x-y\vert}$. Then $d_1$ and $d_2$ are equivalent but $(\mathbb R,d_1)$ cannot be isometric with $(\mathbb R, d_2)$ because $d_2$ is bounded and $d_1$ is not.
To see that "isometric" does not imply "equivalent", take again $X=\mathbb R$. Let $\phi$ be any $discontinuous$ bijection from $\mathbb R$ onto $\mathbb R$ (where $\mathbb R$ has the usual topology). Define $d_1(x,y)=\vert x-y\vert$ and $d_2(x,y)=\vert \phi(x)-\phi(y)\vert$. Then $(\mathbb R, d_1)$ and $(\mathbb R, d_2)$ are isometric by definition, but $d_1$ and $d_2$ are not equivalent.
Etienne
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Consider the standard metric $d_1$ on $\mathbb R$ and let $d_2$ be given by $d_2(x)=\ln (1+ \cdot d_1(x,y))$. These tow metrics give the same topology (the standard topology) on $\mathbb R$, but they are not isometric.
Any isometry between metric spaces, when passing to the topologies induced by the metrics, gives rise to a homeomorphism. Thus, if two metrics on the same space are isometric, then there is an isometry $f:(X,d_1)\to (X,d_2)$, and thus a homeomorphism $X\to X$. If $f$ is the identity, then it implies that the topologies are identical. So, if the identity is an isometry between $(X,d_1)$ and $(X,d_2)$, then the induced topologies are the same.
However, two metric spaces can be isomorphic while the identity is not an isomorphism. For instance, let $g:X\to X$ be a non-identity bijection on a given metric space $(X,d)$. Defining $d':X\times X\to \mathbb R$ by $d'(x,y)=d(g(x),g(y))$ gives a metric space. Clearly, $g:(X,d')\to (X,d)$ is an isometry. But only very rarely will $d$ and $d'$ yield the same topology (they will, of course, give homeomorphic spaces, only the identity need not be a homeomorphism). For a particular example, take $X=\mathbb R$ with the usual metric, and take $g:X\to X$ interchange $0$ and $1$, and leave everything else fixed.
Ittay Weiss
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If $(X,d_1),(X,d_2)$ are isometric metric spaces with identity map then they're equvalent,but otherwise,it's not true in generally,For example $X=\mathbb{R}-0$,$d_2$ is Euclidean metric and $d_1(x,y)=|\frac{1}{x}-\frac{1}{y}|$ now $f:(X,d_1)\to (X,d_2) $s,t $x\mapsto\frac{1}{x}$ is isometry but these are not equivalent because {$n$}$_{n\in\mathbb{N}}$ is Cauchy in $(X,d_1)$.
R Salimi
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