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



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?

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

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


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

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

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    $\begingroup$ Super! thanks...! $\endgroup$ – user444830 Sep 23 '18 at 8:02

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