2

If $X$ is an non-empty set and $d: X \times X \rightarrow \mathbb {R}$ has the following properties

$d(x,y)=0$ if and only if $x=y$

$d(x,y) \leq d(x,z)+\color{red}{d(z,y)}$

Prove that d defines a metric on X.

I need to prove that $d(x,y) \geq 0$

$d(x,y)=d(y,x)$

I know this result.

But the conditions that are set are different, I have tried to do it in an analogous way, but I think that with the conditions that are given it does not meet that it is a metric.

The question falls on the fact that in the statement that I mention you have to d(x,y) $\leq d(x,z)+\color{red}{d(y,z)}$ I would appreciate any hint or if you can help me prove that it is not metric.

Haus
  • 734

1 Answers1

2

Consider $X = \{1, 2\}$ and $d$ defined as $$d(1, 1) = d(2, 2) = 0; \; d(1, 2) = 1; \; d(2, 1) = 2.$$

This clearly satisfies the first condition.
The second condition is trivially satisfied in the following cases:

  1. $z = x,$
  2. $z = y,$
  3. $x = y$.

Therefore, the only case to be checked is when the three are distinct. However, that clearly cannot happen.

Thus, we conclude that the given conditions do not ensure a metric.

Aryaman Maithani
  • 15,771
  • Thank you so much! – Haus Jun 20 '20 at 21:46
  • 2
    In general, this answer shows that a quasi-metric need not be a metric. – Sahiba Arora Jun 20 '20 at 21:48
  • You're welcome! – Aryaman Maithani Jun 20 '20 at 21:48
  • 1
    @SahibaArora: interesting, I'll also point out that I think OP's requirements don't imply that $d$ is a quasi-metric since this $d$ will satisfy OP's conditions even if I change the $1$ to a $-1$. – Aryaman Maithani Jun 20 '20 at 21:59
  • an apology, I have a doubt, in the counterexample you give, what you do is show that it does not meet the second given condition? that is, i.e $d(x,y) \nleq d(x,z)+d(z,y)$ – Haus Jun 21 '20 at 01:39
  • Or what do you mean that the second condition is satisfied with each of the 3 cases it presents? – Haus Jun 21 '20 at 01:51
  • 1
    The counterexample I give DOES satisfy the second condition of yours. However, it does not satisfy the conditions to be a metric since it does not satisfy symmetry. We have $d(1, 2)\neq d(2, 1)$. – Aryaman Maithani Jun 21 '20 at 06:07
  • 2
    To the proposer: If $z=x$ then $d(x,y)\le d(x,x)+d(x,y)=d(x,z)+d(z,y). $ If $z=y$ then $d(x,y)\le d(x,y)+d(y,y)=d(x,z)+d(z,y).$ If $z\ne x$ and $z\ne y$ then $x=y$ (otherwise $x,y,z$ would be $3$ distinct members of $X$) and $d(x,y)=d(x,x)=0\le d(x,z)+d(z,y).$ – DanielWainfleet Jun 21 '20 at 06:16
  • but this would not serve also to exemplify that the example of the link is not a metric? – Haus Jun 21 '20 at 21:40
  • 1
    If you're referring to the quasi-metric link, then yes. I have shown that quasi-metric need not be a metric in general. (And also that the two conditions you gave does not ensure a metric.) – Aryaman Maithani Jun 21 '20 at 21:44