I have begun trying to work through W.R. Scott's Group Theory (which is kind of old, I guess, but hey, Dover is cheap!). And it didn't take me long to get into trouble. Scott says that if we have an equivalence relation $R$ on set $S$, then $S$ is the disjoint union of its equivalence classes w.r.t. $R$.
So let's say $S=\{0,1,2,3\}$ and
$R=\{(0,0),(0,1),(0,2),(1,0),(1,1),(1,2),(2,0),(2,1),(2,2)\}$
Now, they way I see it, $R$ is symmetric, reflexive, and transitive, so it is an equivalence relation. But the union of equivalence classes is $\{0,1,2\}$ so $S$ does not equal the union of equivalence classes.
We might "fix" this by saying that the domain of $R$ is not $S$ but is $\{0,1,2\}$ and we can recover that set through the union of equivalence classes. So $R$ is not really on $S$ but is on this truncated version of $S$. However, the definition of a relation $R$ being on a set $S$ is given as $R\subset S\times S$. $R$ satisfies this definition, so it seems to me that $R$ is indeed on $S$, and we can't fix the problem in this way.
This seems so basic. What am I missing?
$R$ is not reflexive on $S$: it does not contain $(3,3)$.
If you add $(3,3)$ to your set $R$ it will become an equivalence relation on $S$. And in that case you will see that there are exactly two equivalence classes which are disjoint and sum to whole $S$.