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Show that the conditions:

(i) $d(x,y)=0$ iff $x=y(x,y\in X)$ and
(ii) $d(x,z)\le d(x,y)+ d(y,z), \forall x,y,z\in X$

are not sufficient to ensure that the map $d:X\times X \to\mathbb{R}$ is a metric on the set $X$

Can someone provide me the answer please.I am completely stuk onit.

Seirios
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khuku
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    The conditions that the metric be symmetric and the metric be nonnegative are not included. Can you find a counterexample satisfying conditions (i) and (ii) that isn't a metric using this observation? – Amitesh Datta Aug 16 '13 at 07:22
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    @AmiteshDatta: Nonnegativity is in fact a consequence of (ii) since $d(x,x) \leq d(x,x)+d(x,x)$. – Seirios Aug 16 '13 at 07:47
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    That's a neat observation @Seirios; so we know $d(x,x)\geq 0$ for all $x\in X$. How do we know $d(x,y)\geq 0$ when $x,y\in X$ are not necessarily equal? I'm not sure we do; in fact, I can think of a counterexample but perhaps I'm missing something. (E.g., let $X={a,b}$ and $d(a,a)=0=d(b,b)$, $d(a,b)=1$, $d(b,a)=-1$.) If we know that $d$ is symmetric, however, then $d(x,x)\leq d(x,y)+d(y,x)=2d(x,y)$ by (ii) so your observation shows $d(x,y)\geq 0$ for all $x,y\in X$. – Amitesh Datta Aug 16 '13 at 07:55
  • You are right, I wrote too fast... – Seirios Aug 16 '13 at 08:04

3 Answers3

8

By the definition of a metric on the set $X$ it lacks the axiom: $$d(x,y)=d(y,x)\ \forall x,y\in X\tag{*}$$

so we prove by a counterexample that the given axioms in the question don't imply the axiom $(*)$.

Let $X=\{a,b\}$ and $d$ defined by $d(a,a)=d(b,b)=0$ and $d(a,b)=1, d(b,a)=2$ so $d$ verify the axiom (i) and (ii) but not the axiom $(*)$.

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A more natural example: Let $\Gamma$ be a finite weighted oriented multigraph. If $x, y\in \Gamma$ are two vertices, we may define $d(x,y)$ as the length of the shortest oriented path between $x$ and $y$.

  • If all weight is positive then $d(x,y)=0$ iff $x=y$.
  • $d(x,y) \leq d(x,z)+d(z,y)$ for all vertices $x,y,z \in \Gamma$.
  • If $e,f$ are two edges of different weights between two vertices $x,y \in \Gamma$, $d(x,y) \neq d(y,x)$.

For example, such graphs can be used to classify hiking trails with respect to their difficulties. (And clearly, it is easier to go down a hill rather than to climb it, so the "distance" is not symmetric.)

Seirios
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Another example on $\mathbb{R}$: $$\delta(x,y)= \left\{ \begin{array}{cl} |x-y| & \text{if} \ x \leq y \\ \frac{1}{2}|x-y| & \text{if} \ x>y \end{array} \right. .$$

You can verify that $\delta(x,y)=0$ iff $x=y$ and $\delta(x,y) \leq \delta(x,z)+\delta(z,y)$. However, $\delta(x,y)= \delta(y,x)$ iff $x=y$.

Seirios
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