Prove that $d$ and $d^{*}$ are equivalent metrics on $X$. Let $(X,d)$ be a metric space and let $d^{*}(x,y)=\ln{(d(x,y)+1)}$ for all $x,y\in X$.
I need to prove $d^{*}$ is a metric on $X$ equivalent to $d$. But I've no idea of how to do this.
Need help.
 A: Suppose $X=R$ with the Euclidean metric, if $d$ equivalent to $d^*$, $d(0,x)\leq Ad^*(0,x), A>0$ i.e $|x|\leq Aln(|x|+1)$ this is not true for every $x\in R$ since $lim_{x\rightarrow +\infty}{{ln(x+1)}\over x}=0$.
So $d$ is not always equivalent to $d^*$ but they induce the same topology to see this. Let $U$ be an open subset of $(X,d)$ for every $x\in U$ there exists $B_d(x,c)\subset U, B_{d^*}(x,ln(c+1))\subset B_d(x,c)\subset U$. Thus $U$ is open in $(X,d^*)$.
Consider an open subset $V$ of $(X,d^*)$, $x\in V$, $B_{d^*}(x,c)\subset V$, $B_d(x,e^c-1)\subset B_{d^*}(x,c)\subset V$. Thus $V$ is open in $(X,d)$.
A: 1.Prove that $d^*$ is a metric.


*For a sequence $(x_n)_{n\in N}$ in $X, $ and  $x\in X,$  we have $$\lim_{N\to \infty}d(x,x_n)=0 \iff \lim_{n\to \infty}\ln (1+d(x,x_n))=0$$ because the function $f(z)=\ln (1+z)$  , and the inverse function $g(z)=e^z-1, $  are both continuous from $[0,\infty)$ to $[0,\infty).$ 
You can uniquely define any given topology on $X$ in terms of its closure operator, which maps each $Y\subset X$ to $\overline { Y}.$ If a topology on $X$ is generated by a metric $d,$ then $x\in \overline { Y}$ iff there exists a sequence $(x_n)_{n\in N}$ with $\lim_{n\to \infty}d(x,x_n)=0.$ Thus two metrics on X are equivalent if they have the same collection of convergent sequences, because they generate the same closure operator.   Therefore $d^*$ is equivalent to $d.$
(Caution: Equivalent metrics need not have the same set of Cauchy sequences. $d_1$ may be a complete metric, equivalent to an incomplete metric $d_2.$ So if $(x_n)_n$ is a $d_2$-Cauchy sequence with no limit point then $(x_n)_n$ will not be a $d_1$-Cauchy sequence.)         
