Which of the following define a metric on $\mathbb{R}$?

Question

Which of the following define a metric on $\mathbb{R}$?

$d_1(x,y) = \frac{|x|-|y|} {1+|x||y|}$

$d_2(x,y) = \sqrt{|x-y|}$

$d_3(x,y) = |f(x)-g(x)|$ where $f:\mathbb{R}\rightarrow \mathbb{R}$ is strictly monotonic increasing function.

Here is my attempt:

$d_1(x,y)$ satisfies all the three conditions.

$d_2(x,y)$ may fail to satisfy triangle inequality.

$d_3(x,y)$ is not a well defined function.

I am not sure whether i am correct or not? I need a proper justifications.

Thanks for giving me time.

Answer 1

You’re certainly right about $d_3$, since we’re told absolutely nothing about $g$.

To show that $d_2$ may fail to satisfy the triangle inequality, you need to produce an actual example of such a failure. What if $x=0,y=1/2$, and $z=1$?

You need to take another look at $d_1$: what if $|x|<|y|$?

Answer 2

It was recently pointed out in chat that $d(x,y)=\sqrt{|x-y|}$ in fact is a metric. (I am posting this answer mainly so that incorrect - or at least vague - information in the accepted answer is somewhat less prominent.)

See also some other related posts:

• Is $d(x,y) = \sqrt{|x-y|}$ a metric on R?
• Proof of triangle inequality for $d(x,y)=\sqrt{\lvert x-y\rvert}$
• How do I prove that $d(x,y)=(|x-y|)^{\frac{1}{2}}$ is a metric?
• do the following define a metric on $R$?