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The following is sufficient but not necessary condition for topological equivalence:

for each x $\in$ X, there exist positive constants $\alpha$ and $\beta$ such that, for every point y $\in$ X $\alpha d_{1} (x, y) \leq d_{2} (x, y) \leq \beta d_{1} (x, y)$

I am trying to find some exception where above one is true but metrics are not equivalent especially in $R$ domain

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You won’t find such an exception: the condition is sufficient to ensure that the metrics are topologically equivalent. It’s not a necessary condition, however, so it’s possible to have topologically equivalent metrics that do not satisfy the condition. For example,

$$d(x,y)=\frac{|x-y|}{1+|x-y|}$$

is a metric on $\Bbb R$ that generates the usual topology, but there is no $\beta>0$ such that $$|x-y|\le d(x,y)$$ for all $x,y\in\Bbb R$: for any positive integer $n$ we have

$$\frac{|n-0|}{d(n,0)}=\frac{n}{\frac{n}{1+n}}=1+n\;.$$

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  • $\begingroup$ so when that is satisfied we can show that two metrics are equivalent but not the converse $\endgroup$ – Raja Sekar Oct 2 '13 at 7:46
  • $\begingroup$ @RajaSekar: Exactly. $\endgroup$ – Brian M. Scott Oct 2 '13 at 7:48

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