Prove $d$ and $d'=\frac{d}{1+d}$ are equivalent metrics

Question

Suppose $d'(x,y)= \frac{d(x,y)}{1+d(x,y)}$ for $x,y \in X$ and I want to prove $d$ and $d'$ are equivalent metrics on $X$.

I would show $$\lim_{n\rightarrow\infty}d(x_n,x)=0 \quad\Longleftrightarrow\quad \lim_{n\rightarrow\infty}d'(x_n,x)=0.$$

To prove $\lim_{n\rightarrow\infty}d'(x_n,x)=0 \Rightarrow \lim_{n\rightarrow\infty}d(x_n,x)=0$, is it valid to say:

if $0=\lim_{n\rightarrow\infty}d'(x_n,x)=\frac{\lim_{n \rightarrow > \infty}d(x_n,x)}{1+\lim_{n \rightarrow \infty}d(x_n,x)}$,

then $\lim_{n \rightarrow \infty}d(x_n,x)=0$.

If it's invalid, what am I doing wrong?

Thanks!

Answer 1

EDIT

I misunderstood the question - my solution is not valid if you are interested in "topological equivalence", rather it's entirely based around a notion called "strong equivalence". I haven't deleted it however, because it could still be useful.

---

To show two metrics $d, d'$ are strongly equivalent, it doesn't suffice suffice to show that $\lim_{n\rightarrow\infty} d(x_n, x) = 0 \Leftrightarrow \lim_{n\rightarrow\infty} d'(x_n, x) = 0$ for all $x$.

Consider the integers equipped with the usual metric, $d(a,b)=|a-b|$ and the trivial metric, $d'(a,b) = \cases{0, a=b\\ 1, a\neq b}$

Then your above statement with limits is satisfied, because in each case the sequence $x_n$ is eventually constantly equal to $x$. However, $d$ and $d'$ are not equivalent: Let $x \in \mathbb{Z}$. Then there does not exist $c \in \mathbb{R}$ such that $d(x,y)\leq cd'(x,y)$ for all $y\in\mathbb{Z}$.

---

To prove that $d$ and $d'$ are strongly equivalent, you want to:

1. choose an arbitrary $x \in X$.
2. Find $c_1 \in \mathbb{R}$ such that for any $y \in X$, $d(x,y) \leq c_1d'(x,y)$
3. Find $c_2 \in \mathbb{R}$ such that for any $y \in X$, $d'(x,y) \leq c_2d(x,y)$

Answer 2

Ok, if $d(x_n,x)\to 0$, then $$ d'(x_n,x) = \frac{d(x_n,x)}{1+d(x_n,x)}\to \frac 0{1+0} = 0. $$ Conversely, if $d'(x_n,x)\to 0$, then $$ 0\leftarrow \frac{d(x_n,x)}{1+d(x_n,x)} = 1 - \frac 1 {1+d(x_n,x)}. $$ That is, you have $\tfrac{1}{1+d(x_n,x)}\to 1$. Taking reciprocals gives $1+d(x_n,x)\to 1$ and thus $d(x_n,x)\to 0$. You can also do the following: $d' = \tfrac d {1+d}$. Solving for $d$ gives $d = \tfrac{d'}{1-d'}$. Hence, $$ d(x_n,x) = \frac{d'(x_n,x)}{1-d'(x_n,x)}\to\frac 0{1-0} = 0. $$

Answer 3

The function $f(t)=\frac {t}{1+t}$ is strictly increasing for $t\geq 0.$

If $d'(x_n,x)\to 0$ then there are only finitely many $n$ for which $d(x_n,x)> 1/2$. Because $d(x_n,x)> 1/2\implies d'(x_n,x)=f(d(x_n,x))> f(1/2)=1/3$ and we cannot have $d'(x_n,x)> 1/3$ for infinitely many $n.$

So for all but finitely many $n$ we have $d'(x_n,n)=\frac {d(x_n,x)}{1+d(x_n,x)}\geq \frac {d(x_n,x)}{1+1/2}.$

Hence $d(x_n,x)\leq (1+1/2)d'(x_n,x)$ for all but finitely many $n$, so $d(x_n,x)\to 0.$