Two metrics with same topology

Question

Let $M =[1,\infty)$ and let $d_1,d_2:M\times M\rightarrow\mathbb{R}$ be two metrics defined by $$d_1(x,y)=|x-y|,\quad d_2(x,y)=|\frac{1}{x}-\frac{1}{y}|.$$
Prove they induce the same topology.
I already showed that all open sets in $(M,d_1)$ are also open sets in $(M,d_2)$ by showing that all open balls with $d_1$ contain a ball with $d_2$. But i am struggling with the other way.

Answer 1

It is sufficient to show that an open ball around $x$ in one topology contains an open ball around $x$ in the other topology.

Let $r>0$.

Suppose $y \in B_1(x,r)$, then $d_2(x,y) = {|x-y| \over |xy|} \le |x-y| = d_1(x,y) < r$ and so $B_1(x,r) \subset B_2(x,r)$.

Now suppose we are given $B_1(x,r)$, we want to find some $r'$ such that $B_2(x,r') \subset B_1(x,r)$.

Choose $r'=\min({1 \over 2x}, {r \over 2 x^2})$. (Note that if $d_2(x,y) < {1 \over 2x}$ then $y < 2x$.)

Suppose $y \in B_2(x,r')$, then $d_1(x,y) = |x-y| = |xy|{|x-y| \over |xy|} < 2x^2 d_2(x,y) \le r$ and so $y \in B_1(x,r)$.

Note that while the topologies are the same, the Cauchy sequences are different. For example, $x_n = n$ is not Cauchy with $d_1$ but is Cauchy in $d_2$. In particular, $(M,d_1)$ is complete but $(M,d_2)$ is not.

Addendum: To see why $d_2(x,y) < {1 \over 2x}$ implies $y < 2x$: If ${1 \over x} - {1 \over y} \le | {1 \over x} - {1 \over y} | = d_2(x,y) < {1 \over 2x}$, then ${1 \over 2x} < {1 \over y}$ and so $y < 2x$.