Key Concepts and Formulas

• Reflexive Relation: A relation $R$ on a set $A$ is reflexive if $(x, x) \in R$ for all $x \in A$.
• Symmetric Relation: A relation $R$ on a set $A$ is symmetric if whenever $(x, y) \in R$, then $(y, x) \in R$.
• Transitive Relation: A relation $R$ on a set $A$ is transitive if whenever $(x, y) \in R$ and $(y, z) \in R$, then $(x, z) \in R$.
• Equivalence Relation: A relation $R$ on a set $A$ is an equivalence relation if it is reflexive, symmetric, and transitive.
• Partition of a Set: A collection of non-empty subsets $\{A_1, A_2, \dots, A_k\}$ of a set $A$ is a partition if $A_i \cap A_j = \phi$ for $i \ne j$ and $A_1 \cup A_2 \cup \dots \cup A_k = A$.

Step-by-Step Solution

We are given a set $A = A_1 \cup A_2 \cup \dots \cup A_k$, where $A_i \cap A_j = \phi$ for $i \ne j$. This means that the sets $A_i$ form a partition of $A$, and every element $x \in A$ belongs to exactly one $A_i$. The relation $R$ is defined as $R = \{(x, y) : y \in A_i \text{ if and only if } x \in A_i, \text{ for } 1 \le i \le k\}$.

The condition "$y \in A_i \text{ if and only if } x \in A_i, \text{ for } 1 \le i \le k$" means that for a pair $(x, y)$ to be in $R$, the statement $(y \in A_i \iff x \in A_i)$ must be true for every index $i$ from $1$ to $k$.

Step 1: Analyze the condition for $(x, y) \in R$. Let $x \in A$. Since $\{A_1, \dots, A_k\}$ is a partition of $A$, there exists a unique index $p \in \{1, \dots, k\}$ such that $x \in A_p$. For all $j \ne p$, $x \notin A_j$. The condition for $(x, y) \in R$ states: For each $i \in \{1, \dots, k\}$, $(y \in A_i \iff x \in A_i)$ must hold.

Let's consider the implications for a specific $x \in A_p$:

• For $i = p$: $(y \in A_p \iff x \in A_p)$. Since $x \in A_p$ is true, this becomes $(y \in A_p \iff \text{True})$. For this to be true, $y \in A_p$ must be true.
• For $i \ne p$: $(y \in A_i \iff x \in A_i)$. Since $x \notin A_i$ is true (meaning $x \in A_i$ is false), this becomes $(y \in A_i \iff \text{False})$. For this to be true, $y \in A_i$ must be false.

Therefore, for $(x, y) \in R$, if $x \in A_p$, then $y$ must also be in $A_p$. This means that $(x, y) \in R$ if and only if $x$ and $y$ belong to the same subset $A_i$.

Step 2: Check for Reflexivity. To check if $R$ is reflexive, we need to determine if $(x, x) \in R$ for all $x \in A$. Let $x \in A$. Since $\{A_1, \dots, A_k\}$ is a partition, $x$ belongs to exactly one $A_p$. The condition for $(x, x) \in R$ is: for all $i \in \{1, \dots, k\}$, $(x \in A_i \iff x \in A_i)$. This is a tautology for any statement, so it is always true.

• If $i=p$, then $(x \in A_p \iff x \in A_p)$ is (True $\iff$ True), which is True.
• If $i \ne p$, then $(x \in A_i \iff x \in A_i)$ is (False $\iff$ False), which is True. Since the condition holds for all $i$, $(x, x) \in R$. Thus, $R$ is reflexive.

Step 3: Check for Symmetry. To check if $R$ is symmetric, we need to determine if whenever $(x, y) \in R$, then $(y, x) \in R$. Assume $(x, y) \in R$. This means that for all $i \in \{1, \dots, k\}$, $(y \in A_i \iff x \in A_i)$ is true. We need to check if $(y, x) \in R$, which means verifying if for all $i \in \{1, \dots, k\}$, $(x \in A_i \iff y \in A_i)$ is true. The logical biconditional ($\iff$) is commutative. That is, the statement $(P \iff Q)$ is logically equivalent to $(Q \iff P)$. Therefore, if $(y \in A_i \iff x \in A_i)$ is true for all $i$, then $(x \in A_i \iff y \in A_i)$ is also true for all $i$. Thus, $R$ is symmetric.

Step 4: Check for Transitivity. To check if $R$ is transitive, we need to determine if whenever $(x, y) \in R$ and $(y, z) \in R$, then $(x, z) \in R$. Assume $(x, y) \in R$ and $(y, z) \in R$. From $(x, y) \in R$: for all $i \in \{1, \dots, k\}$, $(y \in A_i \iff x \in A_i)$ is true. From $(y, z) \in R$: for all $i \in \{1, \dots, k\}$, $(z \in A_i \iff y \in A_i)$ is true. We need to show that $(x, z) \in R$, which means for all $i \in \{1, \dots, k\}$, $(z \in A_i \iff x \in A_i)$ is true.

Let's consider the statement for a single index $i$. We have the premises:

1. $(y \in A_i \iff x \in A_i)$
2. $(z \in A_i \iff y \in A_i)$

We want to prove $(z \in A_i \iff x \in A_i)$. This is a property of logical implication. If $P \iff Q$ and $Q \iff S$, then $P \iff S$ (transitivity of equivalence). Let $P_i$ be the statement "$x \in A_i$", $Q_i$ be "$y \in A_i$", and $S_i$ be "$z \in A_i$". We are given for all $i$: $(Q_i \iff P_i)$ $(S_i \iff Q_i)$ We need to show for all $i$: $(S_i \iff P_i)$. This follows directly from the transitivity of the logical equivalence relation. Thus, $R$ is transitive.

Step 5: Conclusion on Relation Type. Since $R$ is reflexive, symmetric, and transitive, it is an equivalence relation.

Common Mistakes & Tips

• Misinterpreting "if and only if": The phrase "if and only if" implies a logical biconditional ($\iff$). The condition must hold for all $i$.
• Confusing element membership with set properties: Ensure you correctly apply the partition property that each element belongs to exactly one $A_i$.
• Equivalence relation definition: Remember that an equivalence relation must satisfy all three properties: reflexive, symmetric, and transitive.

Summary

The relation $R$ is defined such that $(x, y) \in R$ if and only if $x$ and $y$ belong to the same subset $A_i$ for some $i$. We have rigorously proven that this relation is reflexive, symmetric, and transitive. Consequently, $R$ is an equivalence relation.

The final answer is \boxed{D}