I am encountering with a question as follows:
Given a set X and $d: X \times X \rightarrow \mathbb{R}$ is a function that satisfies the triangle inequality condition and d(x,y) = d(y,x) condition, but it only follows a flexible version of $d(x,y) \geq 0$ in this case we can have d(x,y) = 0 even when $x \neq y$.
Let $X/\sim$ be the set of equivalence classes for $\sim$ ($x\sim y$ if $d(x,y) = 0$. I claimed that $\sim$ is an equivalence relation.) I then define a function $$d': (X/\sim) \times (X/\sim) \longrightarrow \mathbb{R}$$ as follows: if $E_{0}, E_{1} \in X/\sim$ are equivalence classes, choose $x\in E_{0}, y\in E_{1}$ and set $d'(E_{0}, E_{1}) = d(x,y)$. I am wondering how to prove that $d'$ is well-defined and it is a metric. I found a definition of “well-defined” in group theory, but it seems irrelevant.
Any tip would be great. Thanks!
To say that $d'$ is well-defined means that, if $x\in E_0$ and $y\in E_1$, where $E_0$ and $E_1$ are equivalence classes, then $d(x,y)$ does not depend upon the choice of $x$ and $y$ (within $E_0$ and $E_1$ respectively).
So, take $x'\in E_0$ and $y'\in E_1$. Then\begin{align}d(x',y')&\leqslant d(x',x)+d(x,y)+d(y,y')\\&=d(x,y).\end{align}By the same argument, $d(x,y)\leqslant d(x',y')$. Therefore, $d(x,y)=d(x',y')$.
This proves that your definition of $d'(E_0,E_1)$ makes sense, that is, that $d'$ is well-defined.