How to show whether two metrics generate the same topology In the book metric spaces by Michael O Searcoid, topology of a metric space $X$ is defined as the collection of all open subsets.
Here comes my question. By using this definition, how do we show that two metrics generate the same topology?
The following is the  question: 
Show that the function $d$ define on $\mathbb{R} \times \mathbb{R}$ where $d(a,b)=d(b,a)=2b-2a $ if $0 \leq a \leq b$, $2b-a$ if $a<0 \leq b$, $b-a$ if $a \leq b <0$ is a metric on $\mathbb{R}$ and that it generates the same topology as the Euclidean metric.
I manage to show the first part. But I have no idea on how to show the 2nd part. Can anyone guide me ?
 A: I’ll write $d_E$ for the usual Euclidean metric on $\Bbb R$. It suffices to show two things:


*

*For each $x\in\Bbb R$ and each $\epsilon>0$ there is a $\delta>0$ such that $B_d(x,\delta)\subseteq B_{d_E}(x,\epsilon)$.  

*For each $x\in\Bbb R$ and each $\epsilon>0$ there is a $\delta>0$ such that $B_{d_E}(x,\delta)\subseteq B_d(x,\epsilon)$.


If you’ve learned about bases for topologies, you may recognize this as showing that the the open $d$-balls and the open $d_E$-balls are bases for the same topology on $\Bbb R$.
The reason that this works is simple. Suppose that $U$ is open in the Euclidean topology. Then for each $x\in U$ there is an $\epsilon_x>0$ such that $B_{d_E}(x,\epsilon)\subseteq U$. If you’ve shown (1) above, you know that for each $x\in U$ there is a $\delta_x>0$ such that $B_d(x,\delta_x)\subseteq B_{d_E}(x,\epsilon_x)\subseteq U$, and therefore $U$ is open in the topology generated by $d$ as well.
The argument in the other direction is exactly the same, except that you use (2) instead of (1).
For now I’ll leave it to you to try to prove (1) and (2).
