On $(0,\infty)$, the metrics $d(x,y)=|x-y|+|\frac1x-\frac1y|$ and $d_e(x,y)=|x-y|$ are equivalent.

Question

Problem

Let $Y=(0,\infty)$ and define the metric $d(x,y)=|x-y|+|\displaystyle\frac1x-\displaystyle\frac1y|$ on $Y$. Let $d_e(x,y)=|x-y|$ be the usual Euclidean metric on $Y$, then show that both the metrics $d$ and $d_e$ are topologically equivalent on $Y$.

What I want to show is, any $d$ open ball is $d_e$ open ball and vice versa. Now $B_d(x;r)\subseteq B_{d_e}(x;r)$ proves that any $d_e$ open ball is $d$ open. But for the converse, I could not prove it.

Any hint please!! Or, is there any easy way to look at this problem? Because $d$ is sum of two matrics. Thank you.

Answer 1

Consider the following

Let $f:X\to \Bbb R$ be continuous on a metric space $(X,d)$. Then, $(X,\sigma)$, where $\sigma:X\times X\to [0,\infty)$ defined by $\sigma(x,y)=d(x,y)+\big|f(x)-f(y)\big|$, is topologically equivatent to $(X,d)$.

One side is clear, namely $B_\sigma(a,\epsilon)\subseteq B_d(a,\epsilon)$. To prove the other side, let $a\in X$ and $\epsilon>0$ be given. Using continuity of $f$ at $a$, there is $\delta$ with $\frac{\epsilon}{2}>\delta>0$ such that $d(x,a)<\delta\implies \big|f(x)-f(a)\big|<\frac{\epsilon}{2}$. Now, $B_d(a,\delta)\subseteq B_\sigma(a,\epsilon)$. This is because of the fact that $x\in B_d(a,\delta)\implies \sigma(x,a)=d(x,a)+\big|f(x)-f(a)\big|<\delta+\epsilon/2<\epsilon/2+\epsilon/2=\epsilon.$ So, we are done.