Strongly equivalence and isometric spaces

Question

I have the following definitions

Definition; Two metric spaces $(X,d_1)$ and $(X,d_2)$ are said to be isometric if there is bijective isometry from one space to other.

Definition:Two metric spaces $(X,d_1)$ and $(X,d_2)$ are said to be strongly equivalent if there exists positive real numbers $\alpha$ and $\beta$ such that $$\alpha d_1(x,y) \leq d_2(x,y) \leq \beta d_1(x,y)$$.

My question:As said by Wikipedia that under strong equivalence properties like boundedness ,completeness and all topological properties is preserved.This gives me intuition that notion of strong equivalence of two metrics is same as notion of isometric metric spaces,as isometric spaces has same properties defined from distance function metric on them.But I have no idea,even I don't know whether my claim is true or not.I searched it a lot but didn't find such thing.Is my intuition correct?If not then give some counter example

Thanks!

Answer 1

So as you say, strong equivalence implies that the two metrics define the same topological space. However, they don't define the same metric space.

There is quite a bit of discussion here where you can see what structure is preserved.

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To see something that is not preserved by strong equivalence, consider the following example.

Let $X = \mathbb{R}^2$ and let $d_1(x,y) = \|x-y\|_\infty$ and $d_2(x,y) = \|x-y\|_2$.

Then we can easily see that for all $x,y$ we have that

$$d_(x,y) \le d_2(x,y) \le 2d_1(x,y).$$

So these two metric spaces are strongly equivalent. However they are not isomorphic. Now suppose we are given the point $x = (0,2)$ and we want to find the projection (i.e. the closest point to $x$) onto the set $\{(a,b) : b = 1\} = : S$.

For $d_1$, this projection could be any point in $[-1,1] \times \{ 1\}$.

For $d_2$ the projection is $z = (0,1)$.

Hence, we see have different answers.

However, suppose there was a metric $d_3$ such that $(X,d_1)$ and $(X,d_3)$ are isomorphic. Let $f$ be the isomorphism. Then the projection of $f(x)$ onto $f(S)$ will be exactly $f(z)$. Suppose the projection was instead $\hat{z} \neq f(z)$. Then $\hat{z} \in f(S)$, and $d(f(x), \hat{z}) \le d_3(f(x),f(z))$. Then we have that

$$ d_1(x, f^{-1}(\hat{z})) \le d_1(x,z).$$

However, for specific example we see that this is not possible.