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