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Let $X$ be a set and $d: X \times X \to X$ be a function such that $d(a,b)=0$ if and only if $a=b$.

Suppose further that $d(a,b) ≤ d(z,a)+d(z,b)$ for all $a,b,z \in X$.

Show that $d$ is a metric on $X$.

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Your $<$ must be $\le$. Otherwise your condition can't be true (try $z=a$). –  Robert Israel Dec 2 '12 at 19:52
    
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3 Answers 3

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The first condition of a metric is $d(a,b)\geq 0$ with equality if and only if $a=b$. Obviously that latter portion is satisfied by hypothesis. To show it is greater than zero otherwise, just observe $0=d(b,b)<d(a,b)+d(a,b)$. Thus, the first condition is satisfied.

Next, we want to show $d(a,b)=d(b,a)$. This is clear, though, since $d(a,b)\leq d(b,a)+d(b,b)=d(b,a)$ and vice versa, hence we get equality.

Finally, your last hypothesis is precisely the triangle inequality. Hence, $d$ is a metric.

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Thanks for your answer. Actually, I am not good at this subject, I could not understand how we exactly show d(a,b)≥0 and d(a,b)=d(b,a). –  Ersin Dec 2 '12 at 20:22
    
When we show that $d(a,b)\geq0$, I used the "triangle" inequality of the hypothesis. That is, $$d(b,b)\leq d(a,b)+d(a,b)=2\cdot d(a,b).$$ Dividing by 2 yields $0<d(a,b)$ when $a\neq b$. Now, to show $d(a,b)=d(b,a)$, simply start with $d(a,b)$, and use the triangle inequality from the hypothesis. Then $d(a,b)\leq d(b,a)+d(b,b)=d(b,a)$. On the other hand, $d(b,a)\leq d(a,a)+d(a,b)$. Thus, $d(a,b)\leq d(b,a)\leq d(a,b)$, and so we have equality. Hope this helps! –  Clayton Dec 2 '12 at 20:44
    
Sorry for asking so many questions but I did not understand how we can say d(b,a)≤d(a,a)+d(a,b). Thanks –  Ersin Dec 2 '12 at 21:34
    
And how we say d(b,b)≤d(a,b)+d(a,b). –  Ersin Dec 2 '12 at 21:51
    
That follows from the inequality that we take in the hypothesis. Simply take $z=a$ or $z=b$ depending on the direction. –  Clayton Dec 2 '12 at 22:49

Let $X$ be a set and $d: X \times X \to X$ be a function such that $$d(a,b)=0\text{ if and only if}\;\; a=b,\text{ and}\tag{1}$$ $$d(a,b) ≤ d(z,a)+d(z,b)\forall a,b,z \in X.\tag{2}$$

There's additional criterion that needs to be met for a function $d$ to be a metric on $X$:

  • You must have that $d(a, b) = d(b,a)$ for all $a, b \in X$ (symmetry).

    You can use the two properties you have been given to prove this.

    $d(a,b)\leq d(b,a)+d(b,b)= d(b, a) + 0 = d(b,a)$ and vice versa, hence we get equality.

  • Having proven symmetry, you will then have that

    $d(a,b) \leq d(z,a) + d(z, b) \iff d(a, b) \leq d(a, z) + d(z, b)$.

  • Finally, using the property immediately above, along with the $(1)$, you can establish that for all $a, b\in X$ such that $a\neq b$, we must have $d(a, b) > 0$.

Then you are done.


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Thanks for your answer. Really helped me. –  Ersin Dec 2 '12 at 22:30

This must be the discrete metric. 1) The first condition follows by definition that d(a,b)=0 iff a=b; 2) Symmetry: this is trivial, because if a=b you have d(a,b)=0 and b=a gives you d(b,a)=0; 3) The triangle inequality: from the symmetry you can write d(z,a)=d(a,z). Consider two cases: - if a=b, clear - if a is not equal to b, then either a not equal to z or z not equal to b. There you have $1\leqslant 1$ or $1\leqslant 2$.

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You did not prove symmetry except in the case $a=b$. And where did that $1$ and $2$ come from? –  Robert Israel Dec 2 '12 at 19:55

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