I am trying to prove whether $d_1(x,y)=|x-y|$ and $d_2(x,y)=|\frac{1}{x}-\frac{1}{y}|$ are equivalent metrics or not. Both metrics are defined in the same space $E=\mathbb{R}^+$.
We can check they are not strongly equivalent: $d_2(x,y)=|\frac{1}{x}-\frac{1}{y}|=|\frac{y-x}{xy}|=\frac{|x-y|}{|xy|}\nleq M|x-y|=Md_1(x,y)$, as $|xy|$ can be $<1$ or $>1$ and $M$ has to be a positive constant $\forall x,y\in E$.
I don't succeed proving equivalence. I am trying to prove that, given $r\in\mathbb{R}^+$ and $x\in E,\ \exists s\in\mathbb{R}^+$ such that $B_s(x,d_1)\subseteq B_r(x,d_2)$, where these $B$ are open balls centered in $x$ (and also proving that $B_{s'}(x',d_2)\subseteq B_{r'}(x',d_1)$ given $x'\in E$ and $r'\in\mathbb{R}^+$). Attempt:
$$B_s(x,d_1)=\lbrace y\in\mathbb{R}:d_1(x,y)<s\rbrace=\lbrace y\in\mathbb{R}:|x-y|<s\rbrace=\lbrace y\in\mathbb{R}:\frac{|x-y|}{|xy|}<\frac{s}{|xy|}\rbrace=\lbrace y\in\mathbb{R}:d_2(x,y)<\frac{s}{|xy|}\rbrace$$
But I don't know how to continue. Could you help me? Is there a faster/easier way to prove it? This proof uses $E=(0,1)$. Is that because $d_1$ and $d_2$ are only equivalent in this interval? Thanks in advance!
Hint: $d _1$ and $d_2$ are equivalent iff the following condition holds: For $(x_n) \subset E, x \in E$ we have $d_1(x_n,x) \to 0$ iff $d_1(x_n,x) \to 0$