Equivalence Classes

Question

The scenario is a follows:

I am given a set $X$ along a map $d$ defined as $d:X \times X \to \Bbb R^+$ for all $x,y,z \in X$ with the following properties:

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

The follow equivalence relation is also defined:

$x \tilde{}y \iff d(x,y) =0 $

$\delta$ is defined as such $\delta([x],[y]) = d(x,y)$

I suppose to prove the $\delta$ is well defined; in other words if $[x] =[a]$ and $[y]=[b] $ then $\delta ([x],[y]) = \delta ([a],[b]) $.

I am generally stuck in this case.

I would rather some hints on how to solve this problem in a more general setting

Answer 1

We need to check, that if $[a]=[x]$ and $[b]=[y]$, then $\delta([a],[b])=\delta ([x],[y])$. Using the triangle inequality we obtain: $\delta([a],[b])=d(a,b)\le d(a,y)+d(b,y)=d(a,y)\le d(a,x)+d(x,y)=d(x,y)=\delta([x],[y])$, since by definition $0=d(a,x)=d(b,y)$. The reverse inequality can be shown similarly.

Answer 2

(This answer is based on the old version of the OP's question.)

This is just explicitly writing out my remark in the comments.

As you say, we need to assume $[x]=[a]$ and $[y]=[b]$, and we want to conclude $\delta([x], [y]) = \delta([a], [b])$, that is, that $d(x,y) = d(a, b)$. But by definition, $x \sim a$ means $d(x, a) = 0$, which by the first property, means $x=a$, and similarly we get $y=b$. So in fact, $d(x, y) = d(a, b)$ simply because you have equality of the inputs.