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


1 Answer 1


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,


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


  • $\begingroup$ so when that is satisfied we can show that two metrics are equivalent but not the converse $\endgroup$
    – Raja Sekar
    Oct 2, 2013 at 7:46
  • $\begingroup$ @RajaSekar: Exactly. $\endgroup$ Oct 2, 2013 at 7:48

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