To show that $d $ and $ e$ are equivalent. On the set $X = (0,1]$, consider the usual metric $d(x,y) = |x-y|, (x,y \in X) $ and another  function $e: X\times X \to R$ given by  $e(x,y) = |\frac{1}{x} - \frac{1}{y}|$. Show that $d $ and $ e$ are equivalent.
I am able to show that $e$ is a metric. And for showing they are equivalent we have to show that $B_d (x,r) \subset B_e(x,t)$ and $B_e(x,r) \subset B_d (x,s)$ . And I am able to show that $B_e(x,r) \subset B_d (x,s)$ but unable to show the other part!!
Please Help!!
 A: You need to show that for every $r>0$, there is an $s$ such that $|x-y|<s\to |1/x-1/y|<r$. Let $s=\min(1,xr)\frac{x}2$. Since $s\le\frac x2$ and $|x-y|<s$, you have $y\ge\frac x2$ as well. Then $$\left|\frac1x-\frac1y\right|=\left|\frac{x-y}{xy}\right|\le\frac{|x-y|}{x(x/2)}=\frac2{x^2}|x-y|<\frac2{x^2}\cdot xr\cdot\frac x2=r.$$
(Note that this is the same as the proof that division is continuous, or that $\lim_n\frac{x_n}{y_n}=\frac{\lim_n x_n}{\lim_n y_n}$, if you've seen those proofs before.)
A: To show $B_d (x,r) \subset B_e(x,t)$, we need to show given $x\in (0,1],t>0,\exists r>0$, s.t. $d(x,y)=|x-y|<r\implies e(x,y)=\frac{|x-y|}{|xy|}<t$
As you can see, the key is to obtain a lower bound for the denominator, and since $x$ is fixed, we need to bound $|y|$ away from $0$ using suitable $r$. Then can we? Yes, set $r=\frac{|x|}{2}$, then by triangle inequality,, we have 
$|y|\ge |x|-|x-y|\ge \frac{|x|}{2}$, then we just add another condition to $r$, so that $\frac{|x-y|}{|xy|}<t$ (for example, set $r<\min{(\frac{|x|}{2},\frac{|x|^2}{2}t)}$)
A: Calm down.
For any $x\in (0,1]$ , let $\delta < |x|$ , then $e(x,y) = |\frac{1}{x} - \frac{1}{y}|=|\frac {x-y}{xy}|\le \frac {\delta}{(x-\delta)^2}, \ \forall y\in (x-\delta,x+\delta) $
Let $\delta \rightarrow 0 \Rightarrow \frac {\delta}{(x-\delta)^2}\rightarrow 0$
Thus , $\forall r>0 , \ \exists \delta>0 \ s.t \ \frac {\delta}{(x-\delta)^2}\le r \Rightarrow e(x,y)\le r$
Conclude $B_d (x,\delta) \subset B_e(x,r)$
