Question:

Step 1: Understand what we need to prove. We need to determine whether $A \cap B$ (the intersection of two equivalence relations) is always an equivalence relation. To do this, we must verify that $A \cap B$ satisfies all three properties of an equivalence relation: reflexivity, symmetry, and transitivity. Step 2: Verify reflexivity of $A \cap B$. Since $A$ is an equivalence relation on set $C$, it is reflexive, meaning $(x,x) \in A$ for all $x \in C$. Similarly, since $B$ is an equivalence relation on set $C$, it is also reflexive, meaning $(x,x) \in B$ for all $x \in C$. Therefore, for any $x \in C$, we have: $$(x,x) \in A \text{ and } (x,x) \in B \implies (x,x) \in A \cap B$$ This proves that $A \cap B$ is reflexive. ✓ Step 3: Verify symmetry of $A \cap B$. Assume $(x,y) \in A \cap B$ for some $x, y \in C$. This means: $$(x,y) \in A \text{ and } (x,y) \in B$$ Since $A$ is symmetric (as it is an equivalence relation), $(x,y) \in A$ implies $(y,x) \in A$. Since $B$ is symmetric (as it is an equivalence relation), $(x,y) \in B$ implies $(y,x) \in B$. Therefore: $$(y,x) \in A \text{ and } (y,x) \in B \implies (y,x) \in A \cap B$$ This proves that $A \cap B$ is symmetric. ✓ Step 4: Verify transitivity of $A \cap B$. Assume $(x,y) \in A \cap B$ and $(y,z) \in A \cap B$ for some $x, y, z \in C$. This means: $$(x,y) \in A, \quad (x,y) \in B, \quad (y,z) \in A, \quad (y,z) \in B$$ Since $A$ is transitive (as it is an equivalence relation), from $(x,y) \in A$ and $(y,z) \in A$, we get $(x,z) \in A$. Since $B$ is transitive (as it is an equivalence relation), from $(x,y) \in B$ and $(y,z) \in B$, we get $(x,z) \in B$. Therefore: $$(x,z) \in A \text{ and } (x,z) \in B \implies (x,z) \in A \cap B$$ This proves that $A \cap B$ is transitive. ✓ Step 5: Conclude the result. Since $A \cap B$ satisfies reflexivity, symmetry, and transitivity, it is an equivalence relation on set $C$. Note: The union $A \cup B$ is not necessarily an equivalence relation because it may fail to be transitive. For example, if $(x,y) \in A$ but $(x,y) \notin B$, and $(y,z) \in B$ but $(y,z) \notin A$, then $(x,y) \in A \cup B$ and $(y,z) \in A \cup B$, but $(x,z)$ may not be in either $A$ or $B$, so $(x,z) \notin A \cup B$. **Final Answer:** $A \cap B$ is always an equivalence relation. The correct option is **Option 1**.