# How is it look like an open ball in metric spaced(a,b)=|a−1−b−1| d(a,b)= \left|a^{-1}-b^{-1}\right| on R+\mathbb R^{+}

I try to draw a picture for a ball center at (0,0) with radius 2 but i can get nowhere.

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R+\mathbb R^+ doesn’t include zero, so ab=0ab=0 is not gonna happen, @AbdallahHammam
– Thomas Andrews
2 days ago

Your notation is confusing. R+\mathbb R^+ is the set of positive reals. (0,0)(0,0) is not in the metric space, because elements of the metric space are points on the positive real line.
– Thomas Andrews
2 days ago

B(R+,d)(a;r)B(_{R^{+}},_{d}) ( a;r) @ThomasAndrews you are right there is only line. bu how can it be same as standart metric on R+{R^{+}}
– John dresden
2 days ago

I was wrong about it being the same, but it is very close. (The open sets are the same, but the open balls are slightly different. ) @Johndresden
– Thomas Andrews
2 days ago

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This seems to be a conceptual confusion based on a common confusion about notation.

The confusion is that (0,0)(0,0) is not”a point” in the space. Our “space” is the set R+\mathbb R^+ of (single) positive real number. In this space, it makes no sense to talk about (0,0)(0,0) as a point.

The tricky notation part is there are lots of different means to the notation (x,y)(x,y) going into this question.

There is:

(x,y)(x,y) is a pair of values, like Cartesian coordinates of a point on the plane.
Related, (x,y)(x,y) represents the arguments to the dd function: d(x,y)=|x−1−y−1|d(x,y)=|x^{-1}-y^{-1}|.
(x,y)(x,y) is an open interval – set of all numbers zz such that x14x>\frac{1}{4}.

Note that (R+,d)(\mathbb{R}^+,d) is isometric to (R+,| |)(\mathbb{ R}^+,|\ |)

Proof : Consider f:(R+,d)→(R+,| |)f:(\mathbb{R}^+,d)\rightarrow (\mathbb{ R}^+,|\
|) by
f(a)=a−1 f(a)=a^{-1}

Here |f(a)−f(b)|=|a−1−b−1|=d(a,b) |f(a)-f(b)|=|a^{-1}-b^{-1}|=d(a,b)

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OP probably doesn’t know the meaning of “isometric,” nor why it is important. But yes.
– Thomas Andrews
2 days ago

Yes. Maybe you’re right That is OP may want open ball to know the space (R+,d)(\mathbb{R}^+,d)
– HK Lee
2 days ago

First at all, I asume that your symbol R+\Bbb R^+ is the set {x∈R |x>0}\{x\in\Bbb R\space |x\gt 0\} and your metric space is (R+,d)(\Bbb R^+,d) so neither (0,0)(0,0) nor 00 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 dd of (R+,d)(\Bbb R^+,d)).

Let a∈R+a\in\Bbb R^+; we want to visualize the open ball centered at a(>0)a(\gt 0) and radius 22. By definition we have
B(a;2)={x∈R+:|1a−1x|=|x−a|ax<2}B(a;2)=\left\{x\in\Bbb R^+: \left|\frac 1a-\frac1x\right|=\frac{|x-a|}{ax}\lt2\right\} Let f(x)=|x-a|f(x)=|x-a| and g(x)=2axg(x)=2ax; we need f(x)\lt 2axf(x)\lt 2ax. Note that the slope of f(x)f(x) is -1-1 at the left of aa and 11 at the right of aa while the slope of g(x)g(x) is less than 11 when a\lt \frac 12a\lt \frac 12 and greater than 11 when a\gt \frac 12a\gt \frac 12; it is equal to 11 exactly when a=\frac 12a=\frac 12 It follows that f(x)f(x) and g(x)g(x) have just one intersection point, say x_1x_1, when a\ge \frac 12a\ge \frac 12 and two intersection points, say x_1\lt x_2x_1\lt x_2, when a\lt\frac 12a\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}\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_1x_1 and x_2x_2 as exercise and supplement with two illustrative figures below.