Proving that a metric is non-negative

Question

I wanted to try a problem, where I need to prove that the non-negativity of a metric follows from the following axioms:

For a metric $d$ in some space $X$, we have for $x,y,z\in X$

$$(1)\;\;\;\;d(x,y)=0\;\;\text{iff}\;\;\;x=y$$ $$(2)\;\;\;\;\;\;\;\;\;d(x,y)=d(y,x)$$ $$(3)\;\;\;\;d(x,y)\leq d(x,z)+d(z,y)$$

My question is that, is this a sufficient proof? :

From triangle inequality we have:

$$d(x,y)-d(z,y)\leq d(x,z)$$

By setting $y=x$ it follows from $(1)$ and $(2)$ that:

$$-d(x,z)\leq d(x,z)$$

and this can only be true if $$d(x,z)\geq0$$

http://en.wikipedia.org/wiki/Metric_space#Definition – Surb Feb 03 '15 at 13:51 — Surb, Feb 03 '15 at 13:51
Your argument works. Well done. – Crostul Feb 03 '15 at 13:52 — Crostul, Feb 03 '15 at 13:52
Thank you for your help! Both of you =) – jjepsuomi Feb 03 '15 at 13:53 — jjepsuomi, Feb 03 '15 at 13:53

score 8 · Accepted Answer · answered Feb 03 '15 at 13:51

8

$$0 = d(x,x) \leq d(x,y) + d(y,x) = 2 d(x,y) $$

and so

$$ d(x,y) \geq 0 $$

answered Feb 03 '15 at 13:51

ILoveMath

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Thank you =) appreciate it – jjepsuomi Feb 03 '15 at 13:52

Proving that a metric is non-negative

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