If R and S are two equivalence relations on a set A, then

An equivalence relation on a set A is a relation that satisfies three properties:

Reflexive: Every element relates to itself, $(a,a) \in R$ for all $a \in A$

Symmetric: If $(a,b) \in R$, then $(b,a) \in R$

Transitive: If $(a,b) \in R$ and $(b,c) \in R$, then $(a,c) \in R$

All three properties must hold for a relation to be an equivalence relation.

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Consider $R \cap S$ (the intersection of R and S).

For reflexive: Since both R and S are reflexive, $(a,a)$ is in both R and S. Therefore $(a,a) \in R \cap S$.

For symmetric: If $(a,b) \in R \cap S$, then $(a,b)$ is in both R and S. Since both are symmetric, $(b,a)$ is in both R and S. Therefore $(b,a) \in R \cap S$.

For transitive: If $(a,b) \in R \cap S$ and $(b,c) \in R \cap S$, then both pairs are in R and in S. Since both R and S are transitive, $(a,c)$ is in both R and S. Therefore $(a,c) \in R \cap S$.

Therefore $R \cap S$ is an equivalence relation.

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For any equivalence relation, the inverse relation is also an equivalence relation due to the symmetric property. If $(a,b)$ is in the relation, then $(b,a)$ is already present. Taking the inverse simply reflects the relation.

Since $R \cap S$ is an equivalence relation, $(R \cap S)^{-1}$ is also an equivalence relation.

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Consider $R \cup S$ (the union of R and S).

Transitivity can fail for the union.

Counter-example: Let $(a,b) \in R$ but $(a,b) \notin S$. Let $(b,c) \in S$ but $(b,c) \notin R$. Then both $(a,b)$ and $(b,c)$ are in $R \cup S$. However, $(a,c)$ may not be in either R or S, so $(a,c)$ may not be in $R \cup S$.

Therefore $R \cup S$ is not necessarily an equivalence relation.

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Since $R \cup S$ is not necessarily an equivalence relation, $(R \cup S)^{-1}$ is not necessarily an equivalence relation.

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The intersection of equivalence relations preserves the equivalence properties, while the union does not.

Therefore, $(R \cap S)^{-1}$ is an equivalence relation.