Just 2 relations

The following two facts are direct to prove:

• $\equiv_f \subset A \times A$ defined by $x\equiv_fy \iff f(x)=f(y)$ is an equivalence relation.
• If $R_1,R_2$ are equivalence relations, then $R_1\cap R_2$ is an equivalence relation.

In fact, the first of these facts can be considered the main structure theorem for equivalence relations.

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Then $R = \equiv_f \cap \equiv_g$ is an equivalence relation, but $S = \equiv_f \cup \equiv_g$ isn't necessarily. We can see this by setting $f(x) = |x+1|, g(x) = |x-1| \implies -2 \equiv_f 0 \equiv_g 2 \implies (-2) S 0 S 2$ but $-2 \not\equiv_f 2, \quad -2 \not\equiv_g 2 \implies (-2,2) \not\in S$ so $S$ is not transitive, hence not an equivalence relation.