Let $(X,d)$ a metric space and $\mathcal{R}$ an equivalence relation such that satisfies :
For all $x \in X : [x]$ is closed.
If $[x]\neq [y] $ then for all $a \in [x] : d(a,[y])=d([x],[y])$
Prove that the function : $$ f: \frac{X}{\mathcal{R}} \times \frac{X}{\mathcal{R}} \longrightarrow \mathbb{R} $$ defined by : $f([x],[y])=d([x],[y])$ is a metric in $\frac{X}{\mathcal{R}}$.
I have some problems to prove the triangle inequality. Thanks for read!
Let $x,y,z\in X$ and note that for all $a\in[y]$ we have $$f([x],[z])=\inf_{b\in[z]}d(x,b)\leq \inf_{b\in[z]}d(x,a)+d(a,b)=d(x,a)+f([y],[z])$$ As this holds for all $a\in[y]$ we find $$f([x],[z])\leq \inf_{a\in[y]}d(x,a)+f([y],[z])=f([x],[y])+f([y],[z]).$$