Prove that a metric is always positive.

Question

Given is a set M and a function $d: M \times M \rightarrow \mathbb{R}$ such that :

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

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

To Prove: d is a metric

Thus we need to prove that :

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

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

Any tips on how to go about this?

Also: https://math.stackexchange.com/q/290943/42969, https://math.stackexchange.com/q/3235802/42969, https://math.stackexchange.com/q/1622592/42969, https://math.stackexchange.com/q/1131878/42969. — Martin R, Sep 05 '20 at 11:52

score 2 · Accepted Answer · answered Sep 05 '20 at 11:37

2

Put $z=x$ to get $d(x,y) \leq d(x,x)+d(y,x)=d(y,x)$. Interchange $x$ and $y$ to get the reverse inequality. Hence $d(x,y)=d(y,x)$.

Put $x=y$ to get $0=d(x,x) \leq d(x,z)+d(x,z)=2d(x,z)$. Hence $d(x,z) \geq 0$.

answered Sep 05 '20 at 11:37

Kavi Rama Murthy

1 Answers1