How is it look like an open ball in metric space$ d(a,b)= \left|a^{-1}-b^{-1}\right|$ on $\mathbb R^{+}$ I try to draw a picture for a ball center at (0,0) with radius 2 but i can get nowhere.
 A: This seems to be a conceptual confusion based on a common confusion about notation. 
This answer will address only that, and not the solution.
The confusion is that $(0,0)$ is not"a point" in the space. Our "space" is the set $\mathbb R^+$ of (single) positive real number. In this space, it makes no sense to talk about $(0,0)$ as a point.
The tricky notation part is there are lots of different means to the notation $(x,y)$ going into this question.
There is:


*

*$(x,y)$ is a pair of values, like Cartesian coordinates of a point on the plane.

*Related, $(x,y)$ represents the arguments to the $d$ function: $d(x,y)=|x^{-1}-y^{-1}|$.

*$(x,y)$ is an open interval - set of all numbers $z$ such that $x<z<y$.
It is somewhat important to distinguish between (1) and (2) because the "points" in (1) are pairs $(x,y)$, while the "points" in (2) are individual values, $x$ and $y$. They are, on some technical way, actually the same idea, but it can be confusing when first encountered, and, in metric space discussions, when we talk about "points," we are talking about single elements.
It turns out, you will also need $(3)$ as part of your answer. 
A: Your ball is empty.
For example, the ball $B_o(\frac{1}{2},2)$
of center $\frac{1}{2}$ and radius $2$ is the interval $(\frac{1}{4},+\infty)$.
since we have
$x\in B_o(\frac{1}{2},2) $ iff $d(\frac{1}{2},x)<2$
or
$|2-\frac{1}{x}|<2$  and $x>\frac{1}{4}$.
A: Note that $(\mathbb{R}^+,d)$ is isometric to $(\mathbb{ R}^+,|\ |)$
Proof : Consider $f:(\mathbb{R}^+,d)\rightarrow (\mathbb{ R}^+,|\
|)$ by
$$ f(a)=a^{-1} $$
Here $$ |f(a)-f(b)|=|a^{-1}-b^{-1}|=d(a,b)$$
A: First at all, I asume that your symbol $\Bbb R^+$ is the set $\{x\in\Bbb R\space |x\gt 0\}$ and your metric space is  $(\Bbb R^+,d)$ so neither $(0,0)$ nor $0$ can be center of any non-empty ball. Hence for your question the answer could be the empty ball (recall the empty set is an open in the topology induced by the metric $d$ of $(\Bbb R^+,d)$).  
Let $a\in\Bbb R^+$; we want to visualize the open ball centered at $a(\gt 0)$ and radius $2$. By definition we have
$$B(a;2)=\left\{x\in\Bbb R^+: \left|\frac 1a-\frac1x\right|=\frac{|x-a|}{ax}\lt2\right\}$$
Let $f(x)=|x-a|$ and $g(x)=2ax$; we need $f(x)\lt 2ax$. Note that the slope of $f(x)$ is $-1$ at the left of $a$ and $1$ at the right of $a$ while the slope of $g(x)$ is less than $1$ when $a\lt \frac 12$ and greater than $1$ when $a\gt \frac 12$; it is equal to $1$ exactly when $a=\frac 12$
It follows that $f(x)$ and $g(x)$ have just one intersection point, say $x_1$, when $a\ge \frac 12$ and two intersection points, say $x_1\lt x_2$, when $a\lt\frac 12$. Hence we have
$$\begin{cases}a\ge \frac12\Rightarrow \color{red}{B(a;2)=\{x\gt x_1\}}\\0\lt a\lt\frac 12\Rightarrow \color{red}{B(a;2)=\{x_1\lt x\lt x_2\}}\end{cases}$$
I leave the calculation of $x_1$ and $x_2$ as exercise and supplement with two illustrative figures below.

