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?
[Math] Isometry and equivalence
metric-spaces
Related Solutions
The short answer to "Is there any connection between homeomorphism and equivalence of metric spaces?" is yes. The long answer: any reasonable notion of equivalence of two metrics $d_1$ and $d_2$ can be formulated in terms of the identity map $\mathrm{id}\colon (X,d_1)\to (X,d_2)$. As soon as we distinguish a class of "nice" maps (the class should be a group under composition), we get a notion of equivalence. Some examples, previously mentioned and not:
$\mathrm{id}$ is a homeomorphism
$\mathrm{id}$ is a uniform homeomorphism (i.e., uniformly continuous with a uniformly continuous inverse)
$\mathrm{id}$ is a bilipschitz homeomorphism (i.e., Lipschitz with a Lipschitz inverse)
$\mathrm{id}$ is a quasisymmetric homeomorphism
$\mathrm{id}$ is an isometry
$\mathrm{id}$ is a quasi-isometry
... The list is not exhaustive.
The corresponding notions of equivalence are related by $5\implies 3\implies 2\implies 1$, also $3\implies 4\implies 1$, and $3\implies 6$.
I address some of these issues in this older answer.
But in short: yes $(X,d_1)$ and $(X,d_2)$ are exactly topologically equivalent iff $d_1$ and $d_2$ induce the exact same topology on their set $X$. If the metrics are Lipschitz-equivalent (as they will always be on $\Bbb R^n$, if they come from a norm..., as your examples) then they will be topologically equivalent, but the reverse need not need hold. This is related to the so-called uniform structure on a metric space. Two metrics can be non-equivalent for Cauchy sequences and uniform continuity, but still topologically equivalent.
In topology we typically only care about the topology the metric induces. A metric has other uses, but in pure general topology we choose to ignore them.
Best Answer
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.