Strongly equivalent metrics are equivalent

Question

I have $d,d'$ metrics in X and that they are strongly equivalent. In my case, this means that:

$\exists\alpha,\beta\in\mathbb{R}_{++}$ so that $\alpha d<d'<\beta d$

I want to show that they are equivalent by proving that:

$\forall x\in X,\forall\epsilon>0,\exists\delta>0:B_d(x,\delta)\in B_{d'}(x,\epsilon)$

and vice-versa.

My argument is: take $x\in X,\epsilon>0$ and build the ball $B_{d'}(x,\epsilon)$. If we take the ball $B_{d}(x,\frac{\epsilon}{\alpha})$, then :

$$y\in B_{d}(x,\frac{\epsilon}{\alpha})\Rightarrow d(x,y)<\epsilon/\alpha\Rightarrow\alpha d(x,y)<\epsilon$$

but, using the strongly equivalence property:

$$\alpha d(x,y)<d(x,y)<\epsilon\Rightarrow y\in B_{d'}(x,\epsilon)$$

There's something wrong in this argument. What is it? How can I understand this properly?

Thanks for your time and help!metri

Answer 1

Two metrics are strongly equivalent if they give rise to equivalent notions of strong convergence.

In this sense we want to prove that $$d(x_n,x ) \to 0 \Leftrightarrow d'(x_n,x ) \to 0$$

Assume that for every $B_{d'}(x,\epsilon)$ there is a $\delta$ such that $B_{d}(x,\delta) \subset B_{d'}(x,\epsilon)$

as $x_n \to x$ for large $n$ we have that $x_n \in B_{d}(x,\delta)$ therefore $x_n \in B_{d'}(x,\epsilon)$ therefore $d'(x_n,x) \to 0$.

the other implication is analogous.

What you are trying to prove might be seen more easily as follows:

If $\alpha d <d'<\beta d$ then for every $B_d(x,\epsilon)$ consider $B_{d'}(x, \delta)$

$$d'(y,x)<\delta \Rightarrow d(y,x)< \delta/\alpha $$

So $\delta \leq \epsilon\alpha$ implies that $$B_{d'}(x,\delta) \subset B_{d}(x,\epsilon) $$

the other implication is analogous, take $B_{d'}(x, \delta)$ and consider

$$d(y,x)<\gamma \Rightarrow d'(y,x)< \beta \gamma$$

therefore for $\gamma \leq \frac{\delta}{\beta}$

$$B_{d}(x,\gamma) \subset B_{d'}(x,\delta) \subset B_{d}(x,\epsilon) $$

$$B_{d}(x,\frac{\epsilon}{\alpha \beta}) \subset B_{d'}(x,\frac{\epsilon}{\alpha}) \subset B_{d}(x,\epsilon) $$