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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?

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    An equivalence relation **must** be defined on the whole $\;S\times S\;$, as in reflexivity we require that **for all** $\;x\in S\;,\;\;xRx\;$ . Thus, your relation is *not* an equivalence one. – 2017-01-16
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    +1 for Dover. Since the books are inexpensive you can get several on the same topic. Then you can triangulate, combining different explanations to build your own intuition. – 2017-01-16

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$R$ 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 $S$.

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    OK. Get it now. Thank you. – 2017-01-16