I was reading a book "Algebra" by M. Artin. A lemma was given in that I don't understood. Can you help me?
Lemma 2.7.6: Given an equivalence relation on a set $S$, the subsets of $S$ that are equivalence classes partition $S$.
Provide an example if possible.
Given a set $S$, a partition of $S$ is a collection $\{S_1,\dots ,S_n\}$ of non-empty subsets of $S$ such that $$\bigcup _{i=1}^n S_i=S$$ and $$S_i\cap S_j=\emptyset \text{ for $i\ne j$}$$
The result says that if $R$ is an equivalence relation on $S$ and $S_1,\dots, S_n$ are the equivalence classes of $R$, then $$\bigcup _{i=1}^n S_i=S$$ and $$S_i\cap S_j=\emptyset \text{ for $i\ne j$}$$
For example, consider the relation $a\sim b$ if and only if $a=b$ on the set $\{1, 2\}$. You can easily verify that this relation is reflexive, symmetric and transitive. Thus, it is an equivalence relation. The equivalence classes of this relation are the sets $\{1\}$ and $\{2\}$. As you can see that $$\{1\}\cup \{2\}=\{1, 2\}$$ and $$\{1\}\cap \{2\}=\emptyset$$
The reverse direction of this result is also true. Namely, that if $R=\{S_1,\dots, S_n\}$ is a partition of $S$, then there is an equivalence relation on $S$ whose equivalence classes are the subsets in this partition.
I'll leave the proof of this direction for you to explore!
Edit: Fixed an important part of the theorem that I had mistakingly typed before.