Unique Equivalence Classes

Question

Not sure if this is a suitable question here, but I'm having trouble understanding the intuition behind a theorem I've read in a textbook. So it says the following:

"If $\mathscr R$ is an equivalence relation on a set $A$, then every element $a \in A$ is in exactly one equivalence class. In particular, $a \mathscr R b$ if, and only if, $[a]=[b]$."

I'm having trouble how showing that the "if and only if" statement holding implies that $a$ is in a unique equivalence class. Like I understand proving the "if and only if" part itself, but I don't really know the intuition behind it. I'm guessing it sort of means that either every two equivalence classes are equal or they are disjoint. Could someone please elaborate?

Thanks in advance.

Answer 1

You are correct that any two equivalence classes are equal or disjoint.

Try this for intuition. Suppose you have a bunch of people and each person goes into exactly one room (every room has one or more persons in it). Then person A and person B are in the same room if and only if the set of people in A's room is the same as the set of people in B's room.

Just to be explicit, the underlying set is the set of people. Two people are equivalent (related) if they are in the same room. The equivalence classes are the subsets of people in the various rooms. The equivalence class of a specific person is the set of all people in that person's room (including the person her/him-self).

Answer 2

Assume your if and only if statement .if c is an element of both [a] (all elements equivalent to a) and also an element of [b] then by definition c is equivalent to both a and b so by the definition of equivalence relation,there follows that a and b are equivalent so by the if part follows [a]=[b] . So if [a] and [b] are different (not exactly the same (have some element of one not in the other} then they have no elements in common ,in other words ,disjoint . So you were right ;the iff statement contains this information provided only that you start with an equivalence relation (reflexive,symmetric ,transitive } .Now look back at Paw's nice example . Good question