# Whether 'min' and 'division' are metric

Let $$(X,d_1)$$ and ($$X,d_2)$$ be metric spaces. Whether the following are again metrics on $$X$$ ?

a) $$d(x,y)=\text{min}\;\{d_1(x,y),d_2(x,y)\}$$

b) $$h(x,y)=\Big(\frac{d_1}{d_2}\Big)(x,y)$$ where $$x \neq y$$ and $$h(x,x)=0$$

Actually the answer for the first option is already available in this site. But I mention here is to check my example.

Take $$X=\Bbb{R}$$ and $$d_1(x,y)=\vert x -y\vert$$ and $$d_2(x,y)=\vert x^3-y^3\vert$$ and take $$x=0, y=1/2 ,z=1$$

Now $$d(0,1)=\text{min}\;\{1,1\}=1$$

$$d(0,1/2)=\text{min}\;\{1/2,1/8\}=1/8$$

$$d(1/2,1)=\text{min}\;\{1/2,7/8\}=1/2$$

But $$1=d(0,1) \leq d(0,1/2)+d(1/2,1)=1/8+1/2=0.625$$ does't hold.

Hence $$d$$ is not a metric!

Is this correct? and what about b? Any help?

• Why the downvote? – Theo Bendit Sep 23 '18 at 5:53
• Can you provide a link to the question containing the solution of (a)? (This is unrelated to downvote business) – Devashish Kaushik Sep 23 '18 at 5:53
• Yes! see the question again .I add that link – user444830 Sep 23 '18 at 5:58

a) Yes, your counterexample works perfectly.

b) It is false, for the counterexample let $$d_1$$ be the discrete metric in $$\mathbb{R}$$ (i.e. $$d_1(x,y)=0$$ if $$x=y$$ and $$d_1(x,y)=1$$ if $$x \neq y$$) and $$d_2 = \vert x -y\vert$$ in $$\mathbb{R}$$ . Then

$$h(0,100) + h(100,1) = \frac{d_1(0,100)}{d_2(0,100)} + \frac{d_1(100,1)}{d_2(100,1)} = \frac{1}{100} + \frac{1}{99} < 1$$

and

$$h(0,1) = \frac{d_1(0,1)}{d_2(0,1)} = \frac{1}{1}=1$$

Hence $$h(0,100) + h(100,1) < h(0,1)$$

• Super! thanks...! – user444830 Sep 23 '18 at 8:02