$|.|$ and $d'$ are not equivalent (in metric sense)? Please i need a counter example why $|.|$ and $d'$ are not metricly equivalent where: 
$$d'(x,y)=\left\{
\begin{array}{cc}
  0, & x=y, \\
  |x|+|y|, & x\neq y.
\end{array}\right.$$
Thank you very much .

Hint: This can also help, since metric equivalence implies topological equivalence:
Given any $x \in \mathbb{R}\setminus\{0\}$, let $\varepsilon =\frac{|x|}{2}$, then  the epsilon ball $B_{d'}(x,\varepsilon)$ consists only of $\{x\}$. For, if $y\not= x$ and $y \in B_{d'}(x,\varepsilon)$, we must have $\varepsilon < |x|+|y| < \varepsilon$, a contradiction. For $x=0$, $B_{d'}(0,\varepsilon) = \{x \in \mathbb{R}\ |\ |x|< \varepsilon\} = (-\varepsilon,\varepsilon)$ . Thus, $0$ is the only point that is not open in $(X,d')$.
