Let $A$ be a finite set and $\mathcal R$ an equivalence relation in A. Prove that exists a function $f: A \to \Bbb N$ such that $\forall a, b \in A$ the following is true: $$a \mathcal R b \iff f(a) = f(b)$$
I've already proved that $f(a) = f(b)$ is an equivalence relation. $\mathcal R$ is a subset of the cartesian product $A \times A$, and I know that for $f$ to be a function each element $a \in A$ has to have one and only one image. If any two elements in the domain have the same image I could think of a constant function, that maps every element to the same $n \in \Bbb N$ ... but if this is right, I still don't know how to prove it. So, is ok to think the problem like this? If it is, how would you prove it? I would appreciate your help or tips, thanks!
HINT: Show that the relation has only finitely many equivalence classes, say $C_1,\ldots,C_m$, and use the subscripts on the equivalence classes to define your function $f$.