Finest and Coarsest Equivalences

Question

According to a theorem in the Set Theory book I am reading, we can understand that equivalence relations partitons a set $X$ into distinct equivalence classes, $[x]$. I get that, but one of the problems I have a question on. I think I grasp it, but help would be appreciated. It states.

Each equivalence relation on a set $X$ defines a partition. What equivalence yields the finest partition? The coarsest?

So I was thinking of coarse as broad, so the equivalence congruence modulo $2$ struck me since this partitions the integers into $[0]$ and $[1]$. However, the set here is only $\mathbb{Z}$. Now the finest could be the equivalence $R$ where $(x,y)R(u,v)$ if $xv=yu$, since there are infinite equivalence classes then (as elements of $\mathbb{Q}$ such that $\gcd(a,b)=1$, where $a,b\in\mathbb{Z}$.

Is this the right approach? Is this the correct intuition of "coarse" and "fine" equivalences and are there coarser and finer equivalences? Is it subjective to the particular set being partitioned?

Answer 1

Hint: The partition $\{X\}$ is even coarser ...

And the one that partitions $X$ into singletons is even finer.

Your book really ought to contain a precise definition of what "coarser" and "finer" means in this context.

Answer 2

If $X$ is a set and $P_1$ and $P_2$ are partitions of $X$, then we say $P_1$ is coarser than $P_2$ (and $P_2$ is finer than $P_1$) if for all $S \in P_2$, there exists a set $T \in P_1$ such that $S \subseteq T$.

A partition of $X$ has size at least $1$ and at most $|X|$. I suggest you see if these are possible to achieve as they would be the coarsest and finest possible automatically.