Let a set A = A_{1} cup A_{2} cup ldots cup A | vedclass

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(D) The relation $R$ is defined as $(x, y) \in R$ if and only if $x$ and $y$ belong to the same subset $A_{i}$.$1$. Reflexivity: For any $x \in A$,$x$

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an equivalence relation

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(D) The relation $R$ is defined as $(x, y) \in R$ if and only if $x$ and $y$ belong to the same subset $A_{i}$.$1$. Reflexivity: For any $x \in A$,$x$ belongs to some $A_{i}$. Since $x \in A_{i} \iff x \in A_{i}$,$(x, x) \in R$. Thus,$R$ is reflexive.$2$. Symmetry: If $(x, y) \in R$,then $x$ and $y$ belong to the same $A_{i}$. This implies $y$ and $x$ belong to the same $A_{i}$,so $(y, x) \in R$. Thus,$R$ is symmetric.$3$. Transitivity: If $(x, y) \in R$ and $(y, z) \in R$,then $x, y \in A_{i}$ and $y, z \in A_{j}$. Since $A_{i} \cap A_{j} = \phi$ for $i 
eq j$,we must have $i = j$. Therefore,$x, z \in A_{i}$,which means $(x, z) \in R$. Thus,$R$ is transitive.Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation.

Let a set $A = A_{1} \cup A_{2} \cup \ldots \cup A_{k}$,where $A_{i} \cap A_{j} = \phi$ for $i \neq j$ and $1 \leq i, j \leq k$. Define the relation $R$ from $A$ to $A$ by $R = \{(x, y) : y \in A_{i} \text{ if and only if } x \in A_{i}, 1 \leq i \leq k\}$. Then,$R$ is

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