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.

  • 1
    $\begingroup$ Note that $d_{1}(-1,1)=0$. $\endgroup$
    – T. Eskin
    May 17, 2012 at 10:55
  • $\begingroup$ @ThomasE. Great comment. Simple and elegant. $\endgroup$
    – Srijan
    May 17, 2012 at 11:20
  • $\begingroup$ @srijan What is $g$?$g$ is not in the question. Isn't it $d(x,y)=|f(x)-f(y)|?$. isn't it a question of NBHM? $\endgroup$
    – user464147
    Oct 25, 2017 at 13:34

2 Answers 2


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 $|x|<|y|$?

  • $\begingroup$ if we have $d_1(x,y) = \frac{|x-y|} {1+|x||y|}$ , can we say that it will form metric space? $\endgroup$
    – Srijan
    May 17, 2012 at 10:16
  • 1
    $\begingroup$ @srijan: What happens if $x=0,y=1,z=2$? Does the triangle inequality hold for all permutations of those three points? $\endgroup$ May 17, 2012 at 10:22
  • $\begingroup$ Certainly it will not satisfy triangle inequality for $x = 0$, $y = 2$, $z = 1$. Heartily thanks to you sir. $\endgroup$
    – Srijan
    May 17, 2012 at 10:27
  • $\begingroup$ One last question sir: What if g is also strictly monotonically increasing function? $\endgroup$
    – Srijan
    May 17, 2012 at 10:30
  • 1
    $\begingroup$ @srijan: And $d_3(x,y)=|f(x)-g(y)|$? It won’t satisfy $d_3(x,x)=0$ for all $x$ unless $f=g$. $\endgroup$ May 17, 2012 at 10:32

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.)

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