Homeomorphism and equivalent metrics

Question

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?

Answer 1

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.

Answer 2

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 $\alpha d(x,y)<|x-y|<\beta d(x,y)$.