# Homeomorphism and equivalent metrics

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?

• For norms the situation is indeed different: the boundedness of the identity in both directions will give you such constants as you require. – Henno Brandsma Mar 19 at 22:11

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.
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 $$\alpha d(x,y)<|x-y|<\beta d(x,y)$$.
• 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? – Mike V.D.C. Mar 21 at 13:52
• the two metrics induces the same topology, but they are not equivalent (thers is NO constant $\alpha$ and $\beta$...) – user126154 Mar 22 at 13:03