Adjusted Axioms / Postulates for a metric space. Let $D$ be a function defined for all $a$ and $b$ in the metric space $M$
I. $D(a,a)=0 \, \forall a$
II. $D(a,b)\geq0$ [the only adjusted one, CRUCIAL]
III. $D(a,b)=D(b,a)$
IV. The Triangle inequality
Question: Define $A \sim B$ to mean that $D(a,b)=0$. Prove equivalence relation for $\sim. $
How does one do it?
My attempt:
a) $a \sim a\,$ mean that $D(a,a)=0$ which is true according to axiom I.
b) According to axiom III we have that $D(a,b)=D(b,a)$. Since $a \sim b$ mean that $D(a,b)=0$ that means that also $D(b,a)=0$ therefore $b \sim a$.
c) How to show transitivity, here is where I am stuck at the moment.
If $D(a,b)=D(b,c)=0$ then observe by triangle inequality that $D(c,a) \leq D(a,b) + D(b,c) = 0 \implies D(c,a) = 0.$