Exercise 3.2

Proof. Define $C$, $A$, and $A_0$ as above and suppose that $C$ is an equivalence relation. Let $C_0=C\cap (A_0\times A_0)$ be the restriction of $C$ to $A_0$, noting that this is in fact a relation on $A_0$ since clearly $C_0\subset A_0\times A_0$. Now we show that $C_0$ satisfies the three properties of an equivalence relation.

• (Reflexivity) Consider any $a\in A_0$ so that of course $(a,a)\in A_0\times A_0$. Since $A_0\subset A$ we also have that $a\in A$. Hence $\mathit{aCa}$ since $C$ is an equivalence relation on $A$ and is therefore reflexive. Thus $(a,a)\in C\cap (A_0\times A_0)=C_0$, which shows that $a{C}_{0}a$ so that $C_0$ is reflexive since $a$ was arbitrary.
• (Symmetry) Suppose that $a,b\in A_0$ and that $a{C}_{0}b$. Then of course $(b,a)\in A_0\times A_0$ and $\mathit{bCa}$ since $C_0\subset C$. From this it follows that $(b,a)\in C\cap (A_0\times A_0)=C_0$ so that $b{C}_{0}a$. This of course shows that $C_0$ is symmetric.
• (Transitivity) Now consider $a,b,c\in A_0$ and suppose that both $a{C}_{0}b$ and $b{C}_{0}c$. Then we have $\mathit{aCb}$ and $\mathit{bCc}$ since $C_0\subset C$. Since $C$ is an equivalence relation and therefore transitive, it follows that $\mathit{aCc}$, and since also clearly $(a,c)\in A_0\times A_0$, we have $(a,c)\in C\cap (A_0\times A_0)=C_0$ so that $a{C}_{0}c$. This shows that $C_0$ is transitive.