Confusion on why 2 equivalence classes are either equal or disjoint

Question

So I've just started trying to teach myself some topology and in the book I'm reading there is a proof that 2 equivalence classes are either equal or disjoint.

However I'm a bit confused on why the author just randomly states any 2 equivalence classes surely it should be any 2 equivalence classes defined by the same equivalence relation or is it that any 2 equivalence classes regardless whether or not the equivalence relations on them is equal or not are either equal or disjoint?

Thanks in advance.

Answer 1

Yes, the equivalence classes must be defined by the same equivalence relation. Otherwise the statement is not true. For example,take the set $X=\{1,2,3\}$ ande define these equivalence relations:

1. for all $x,y\in X$, set $x$ and $y$ equivalent
2. for all $x\in X$, $x$ is equivalent only with itself and with no other elements.

In the first case there is only one equivalence class, namely the whole $X$. In the second case there are three equivalence classes, $[1]$,$[2]$ and $[3]$. As you can see the classes of the second equivalence relation are not dijoint from the only class of the first equivalence relation.

Answer 2

Yes, the discussion is about one fixed equivalence relation $R$ on $X$, say.

If two classes $[x]_R$ and $[x']_R$ of $R$ intersect, $xRx'$ (transitivity via the intersection element) and so the classes are actually equal to each other.

We rarely mix equivalence relations; there is usually one under consideration, and we discuss its classes.