Show that d is a metric if it fulfills triangle inequality and $d(x,y)=0\iff x=y$

Question

I have the following exercise:

$M$ is a non empty set and $d: M\times M \to \mathbb R$ and application such that:

a) $d(x,y)=0\iff x=y$

b) $d(x,z) \le d(x,y)+d(y,z)$

show that $d$ is a metric

Aren't those 2 of the properties for a metric space? Also, wouldn't $b)$ imply that:

$$d(x,x)\le d(x,y)+d(y,x) \implies 0 \le d(x,y)+d(y,x)$$

which if we accept that $d(x,y) = d(y,x)$:

$$0 \le 2d(x,y) \implies d(x,y) \ge 0$$ ?

So, wouldn't it be necessary for a metric space to have only:

$$d(x,y)\ge 0$$ $$d(x,z)\le d(x,y)+d(y,z)$$ $$d(x,y)=0\iff x=y$$ ?

What am I not getting here?

UPDATE: the right question is:

$M$ is a non empty set and $d: M\times M \to \mathbb R$ and application such that:

a) $d(x,y)=0\iff x=y$

b) $d(x,z) \le d(x,y)+d(z,y)$

show that $d$ is a metric

I don't see the exercise... What is the claim? The problem in what you are doing above is "if we accept that". — Friedrich Philipp, Mar 11 '16 at 02:25
Ok.Obviously, you have to show that $d(x,y)\ge 0$ and $d(x,y) = d(y,x)$ holds for all $x,y$. — Friedrich Philipp, Mar 11 '16 at 02:30
@FriedrichPhilipp but $d(x,y)\ge 0$ can be shown if we accept $d(x,y) = d(y,x)$, so why do we need this assumption for metric spaces? — Guerlando OCs, Mar 11 '16 at 02:31
I don't understand this question. The task here is to show that you don't need any of these two assumptions. — Friedrich Philipp, Mar 11 '16 at 02:35
@FriedrichPhilipp why we need to assume that $d(x,y)\ge 0$, in a metric space, if the triangular inequality + the assumption that $d(x,y)=d(y,x)$ implies that $d(x,y)\ge 0$, as I did in my question? — Guerlando OCs, Mar 11 '16 at 02:36
These two properties arent enough to show that $d$ is a metric, you need $d\ge 0$ and $d(x,y)=d(y,x)$. The exercise is senseless. — Masacroso, Mar 11 '16 at 02:38
@Masacroso That's what I thought, but I couldn't come up with a counterexample. — Friedrich Philipp, Mar 11 '16 at 02:39
A huge load of counterexamples are here: http://math.stackexchange.com/questions/23390/examples-of-non-symmetric-distances — Sarvesh Ravichandran Iyer, Mar 11 '16 at 02:40
@FriedrichPhilipp try just two points, $M={a,b}$, $d(a,b)=1$, $d(b,a)=-1$. — Mirko, Mar 11 '16 at 02:44
@FriedrichPhilipp maybe metric based in a p-norm with $p\in(0,1)$ with a minus multiplying the result. Then the 2 statements hold (subadditivity and 0 if is the same number). — Masacroso, Mar 11 '16 at 02:52
What does "and application" mean? And why wouldn't my previous comment provide a counterexample? — Mirko, Mar 11 '16 at 02:58

score 4 · Accepted Answer · answered Mar 11 '16 at 05:39

So you have a function $d: M\times M \to \mathbb R$ satisfying -

a) $d(x,y)=0\iff x=y$

b) $d(x,z) \le d(x,y)+d(z,y)$

As you noted, you just have to show that it takes non-negative values and that the symmetry property is satisfied. Also you need to show triangle inequality (which is obvious from (b) once you've shown symmetry)

To show non-negativity (as you have done) - $$d(x,x)\le d(x,y)+d(x,y) \implies 0 \le 2d(x,y)\implies d(x,y)\ge0$$

To show symmetry we play around with (b)-

Taking $y=x$ in (b) we get, $$d(x,z)\le d(x,x)+d(z,x) \implies d(x,z) \le d(z,x)$$ Similarly interchanging $x$ and $z$ and taking $y=z$ we get, $$d(z,x)\le d(z,z)+d(x,z) \implies d(z,x) \le d(x,z)$$ So that $$d(x,z)=d(z,x)$$

Show that d is a metric if it fulfills triangle inequality and $d(x,y)=0\iff x=y$

1 Answers1