$D(x,y)=|1/x-1/y|$ equivalent to the standard metric $d$ on $(0,1]$?

Question

Is the metric $D(x,y)=|1/x-1/y|$ equivalent to the standard metric $d$ on $(0,1]$?

If so,how?

I know that I have to show for $\epsilon >0$ there exists $\delta >0$ such that $B_D (x, \delta) \subset B_d (x, \epsilon)$.

Also,what about the metric $P(x,y)=|x/({1+|x|})-y/({1+|y|})|$? Is it equivalent to the usual metric on $R$?

Answer 1

Yes, they are equivalent because $f: ((0,1],D) \to ([1, \infty),d)$ defined by $f(x) = \frac{1}{x}$ is an isometry:

$$d(f(x), f(y)) = |\frac{1}{x} - \frac{1}{y}| = D(x,y)$$

and $((0,1],d)$ and $([1,\infty),d)$ are homeomorphic, vis that same $f$.

Answer 2

Two metrics are equivalent iff they have the same convergent sequences with the same limits. In the first case this condition is satisfied so the metrics are equivalent. In the case of $P(x,y)$ note that $P(x_n,x) \to 0$ implies $\frac {x_n} {1+|x_n|} \to \frac x {1+|x|}$. Taking absolute value on both sides we get $\frac {|x_n|} {1+|x_n|} \to \frac {|x|} {1+|x|}$. This implies $|x_n| \to |x|$. Going back to $\frac {x_n} {1+|x_n|} \to \frac x {1+|x|}$ we see that $x_n \to x$. Converse part is straightforward. So $P$ is also equivalen to to the standard metric.