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Let $d_1$ and $d_2$ be two metrics on a space $M$ such that the metric spaces $(M, d_1)$ and $(M, d_2)$ are homeomorphic to each other. I know that if the identity map is continuous, then metrics are equivalent.

However, I am not able to go beyond this. In other words, I can neither prove nor able to produce a counterexample to the statement that

homeomorphism between the spaces implies the metrics are equivalent.

My definition of equivalence of metrics is there exists $\alpha,\beta$ such that for every $x,y\in M$, $$\alpha d_1(x,y)\leq d_2(x,y)\leq \beta d_1(x,y).$$ As the spaces are homeomorphic, an open set in one space is open in other. But this may not imply that a $\epsilon$-ball in one space is a $\delta$-ball in other.

Can someone help me the clarifying this?

-- Mike

P.S.:

Does the situation become different in the case of normed linear spaces?

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    $\begingroup$ For norms the situation is indeed different: the boundedness of the identity in both directions will give you such constants as you require. $\endgroup$ – Henno Brandsma Mar 19 at 22:11
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No, it's not true. As your definition of equivalence between metrics implies $(M,d_1)$ is bounded if and only if $(M,d_2)$ is bounded, but boundedness is not a topological property, i.e. it's not preserved by homeomorphisms.

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A nice example is the square root metric on $\mathbb R$.

Define $d(x,y)=\sqrt{|x-y|}$.

This is a metric on $\mathbb R$ and induces the Euclidean topology. But there exists no constants $\alpha,\beta>0$ such that $\alpha d(x,y)<|x-y|<\beta d(x,y)$.

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  • $\begingroup$ Sorry I am slightly confused. Does this give a homeomorphism or the metrics are equivalent? (I am not able to show either way...) My attempt: I showed that Identity map is continuous (by showing inverse image of $B_\epsilon(x)$ is open), so the metrics must be equivalent. But I am not able to find $\alpha,\beta$. Am I wrong in showing this? Am I missing something? $\endgroup$ – Mike V.D.C. Mar 21 at 13:52
  • $\begingroup$ the two metrics induces the same topology, but they are not equivalent (thers is NO constant $\alpha$ and $\beta$...) $\endgroup$ – user126154 Mar 22 at 13:03

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